SCHOOL OF MATHEMATICS

MATHEMATICS FOR PART I ENGINEERING

Self-paced Course

MODULE 2 TRIGONOMETRY

Module Topics

1. The measurement of angles and conversion between degrees and radians

2. The elementary trigonometric ratios and relations between them

3. The solution of general triangles

4. Trigonometric ratios for various angles and graphs of the elementary trigonometric functions

5. Other trigonometric ratios and relationships between them

6. Various useful formulae involving multiple angles

7. Trigonometric equations

This module is again self-contained. It is another long module but it should be mainly revision for you.

Work Scheme

Study the following sections, read carefully the worked Examples and do the stated Exercises. Solutions tothe Exercises are given towards the end of this Module, starting on p.21.

1. The measurement of angles Angles are usually measured either in degrees or in radians. Thereare, of course, 360 degrees (360◦) in a complete revolution, 180◦ in a ‘straight line’ angle and 90◦ in aright-angle.

By definition, if the angle θ in figure 1 is measured in radians,

......................................................................................................................................................................................................................

...............

...............

r L

θ

Figure 1

.......

.......

.......

.......................................................................

.......................

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................................................

.......

.............................

.......

.......

.......

...........

................................

then θ =Length of arc

radius=L

r,

a ratio which is independent of the units in which both L and r are measured.

Since the length of the complete circumference of a circle of radius r is 2πr, in a complete revolution thereare 2πr/r = 2π radians.

Conversion from degrees to radians and vice-versa is straightforward since 360◦ is equivalent to 2π radians:

To convert from degrees to radians, multiply byπ

180.

To convert from radians to degrees, multiply by180

π.

– 1 –

In trigonometry it is usually most convenient to work in radians and, therefore, you may assume that allangles are measured in radians unless stated otherwise. If angles are measured in degrees then the usualnotation will be used (e.g. 45◦).

When using a pocket calculator, you must be very careful that it is operating in the correct—i.e. radian—mode unless you specifically intend to use degrees. Remember, though, that an accuracy of four decimalplaces in describing an angle in radians is equivalent to an accuracy of only two or three places in degrees.The answers to many problems in this Module are given in both radians and degrees. These answers are(hopefully!) accurate to four places. However, if you convert one answer directly to the other by using theappropriate scale factor, because of rounding errors in the calculation you may well find an apparent errorin the last one or two decimal places.

Example 1. Convert 60◦ to radians.

60◦ is 60× π

180=π

3radians.

Example 2. Convertπ

6radians to degrees.

π

6radians is

π

6× 180

π

◦= 30◦.

One or two frequently used conversions are worth committing to memory:

30◦ =π

6, 45◦ =

π

4, 60◦ =

π

3, 90◦ =

π

2, 180◦ = π , 360◦ = 2π .

For other cases, use your pocket calculator (or tables) if necessary.

***Do Exercise 1. Express the following in degrees: (i) π/4, (ii) 3π/2, (iii) 7π/3.

(N.B. In all the Exercises in this module give numerical answers correct to 4 decimal places (i.e. 4 d.pl.),using degrees or radians as appropriate.)

***Do Exercise 2. Express the following in radians: (i) 18◦, (ii) 35◦, (iii) 350◦, (iv) 420◦.

2. The three elementary trigonometric ratios In a right-angled triangle, the sine, cosine and tan-gent of an acute angle are defined to be:

Sine of angle =side oppositehypotenuse

, Cosine of angle =adjacent sidehypotenuse

Tangent of angle =side oppositeadjacent side

.

The notations used for the three ratios are respectively sin, cos and tan. Thus, in the right-angled triangleshown in figure 2 (the side labelled c, opposite the right-angle is, of course, the hypotenuse):

sin θ =b

c, cos θ =

a

c, tan θ =

b

a.

.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. .......

.......

.......

.......................

a

bc

θ

Figure 2

.......

...............................

– 2 –

Once again, assuming of course that all lengths are measured in the same units, these ratios are independentof which units are used. This means that, given a right-angled triangle, if the length of any side and thesize of a second angle are known or if the lengths of any two sides are known then the triangle can be‘solved’—that is to say all other sides and angles of the triangle can be found.

Example 3. Solve the triangle shown in figure 2 if a = 2 cm and θ = π/6.

Since tan θ = b/a, cos θ = a/c, it follows that b = a tan θ and c = a/ cos θ and hence, putting in thegiven values for a and θ, we see that

b = 2 tan(π

6

)= 1.1547 cm and c =

2

cos(π/6)= 2.3094 cm .

[Note that we do not need trigonometry to calculate the third angle—the sum of the angles of atriangle is π and so the value of the third angle is π/3. Also, having found b we could have usedPythagoras’ theorem to give us c.]

Example 4. Solve the triangle shown in figure 2 if a = 2 cm and b = 1 cm.

Firstly we use Pythagoras’ theorem to find c:

c =√a2 + b2 =

√22 + 12 =

√5 = 2.2361 cm .

Next, since

tan θ =b

a= .5 ,

the inverse tangent button on your pocket calculator gives θ = .4636 or 26.5651◦. The remainingangle is then π − ( 1

2π + .4636) = 1.1072 or 180◦ − (90◦ + 26.5651◦) = 63.4349◦.

3. Important relations between sin, cos and tan Referring again to figure 2 you can see that, byPythagoras’ theorem,

c2 = a2 + b2 .

However, a = c cos θ and b = c sin θ and hence

c2 = c2 cos2 θ + c2 sin2 θ ,

from which it follows thatcos2 θ + sin2 θ = 1 .

Notice the notation carefully. We write, for example, cos2 θ to mean (cos θ)2. [The quantity cos θ2 has nomeaning since it could represent cos(θ2) or (cos θ)2].

Example 5. Verify the formula cos2 θ + sin2 θ = 1 for the angle θ in Example 4.

Since θ = .4636, your pocket calculators tell you that cos2 θ = .8, whilst sin2 θ = .2. Hence cos2 θ +sin2 θ = 1.

Another relationship between cosine and sine is obvious from figure 2. The angles between a and c andbetween b and c add to π/2 since the third angle of the triangle is π/2 and we know that the angles of atriangle add to π. It follows immediately that

cos(π

2− θ)

= sin θ ,

sin(π

2− θ)

= cos θ .

– 3 –

In addition it is easily shown that

tan θ =sin θ

cos θ,

since

tan θ =b

a=b/c

a/c=

sin θ

cos θ.

You can check for yourself that the above three formulae involving θ are true for the numbers given inExample 4.

The four major results in this section are identities—that is to say, they hold for all values of θ. They areimportant and you should be sure to memorise them. They are repeated as numbers 4, 5 and 7 of the ‘usefulresults’ at the end of this module.

4. The solution of general triangles If you are given any triangle together with any one of the fol-lowing three pieces of information:

(a) the lengths of all three sides

(b) the lengths of two sides and the size of any angle

(c) the length of one side and the size of any two angles

you can still ‘solve’ the triangle completely although in the second case, unless the given angle lies betweenthe two sides, the solution may not be unique. To do this you can use two rules called respectively the‘sine’ rule and the ‘cosine’ rule. In the notation used in figure 3 (which is the conventional notation with aopposite A etc.), the ‘sine’ rule is written

a

sinA=

b

sinB=

c

sinC

and the ‘cosine’ rule gives the three formulae

a2 = b2 + c2 − 2bc cosA, b2 = c2 + a2 − 2ca cosB, c2 = a2 + b2 − 2ab cosC.

.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..................................

..................................

..................................

..................................

..................................

..................................

..................................

..................................

..................................

............................................................................................................................................................................B C

A

a

bc

Figure 3

The table below shows you which formula you need for which situation.

Facts known Rule to use

3 sides cosine2 sides and the angle between

2 sides and either of the other angles sine1 side and any 2 angles

The sine and cosine rules appear on the Formula Sheet (and in the Data Book) but it is useful to rememberthem. You do not need to know the proofs of the two rules but they are included below since they are verystraightforward.

– 4 –

........................................................................................................................................................................................................................................................................................................................................................................................

....................

....................

....................

....................

....................

....................

....................

....................

....................

....................

....................

....................

......................................................................................................................................................................................................................................................

B C

A

F

a

bhc

Figure 4

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

In figure 4, you can see that the height, h, of the triangle is given equivalently by h = c sinB and h = b sinC.Hence c sinB = b sinC and

b

sinB=

c

sinC.

The remaining part of the sine rule is proved similarly or by symmetry.

To prove the cosine rule we see, again looking at figure 4,

c2 = h2 + (a− CF )2 = b2 sin2 C + (a− b cosC)2

= b2 sin2 C + a2 + b2 cos2 C − 2ab cosC .

However, cos2 C + sin2 C = 1, and hence c2 = a2 + b2 − 2ab cosC.

Once again, the other two results follow similarly or by symmetry.

Example 7. Solve the triangle ABC if a = 12, b = 8, c = 10.

Since all three sides but no angles are known, we must use a cosine formula.c2 = a2 + b2 − 2ab cosC tells us that

cosC =a2 + b2 − c2

2ab

=144 + 64− 100

2× 12× 8

=108

192= 0.5625 .

Hence, using a calculator, C = .9734, (= 55.7711◦).

We could find the other angles in a similar way, but we now know one angle and so a simpler alternativeis to use the sine formula. Since

a

sinA=

c

sinC,

sinA =a sinC

c

=12× .8268

10= .9922 .

Hence A = 1.4455, (= 82.8192◦).

To find the remaining angle we could use the cosine or the sine formula, but obviously the easiest way isto remember that the angles of a triangle add to π, so that B = π−A−C = π−1.4455−.9734 = .7227,(= 41.4097◦). Notice that these results have been obtained retaining 10 digits throughout on a pocketcalculator. However, at each stage they have been stated to only four places. If you use the four placeanswers (instead of the full 10 place ones) to perform your subsequent calculations, you will obtainan answer which is certainly not accurate to four places. Try it and see!

– 5 –

Example 8. Solve the triangle ABC if b = 2 cm, c = 1 cm, C = 20◦.

Given two sides and non-included angle, we must use the sine formula. In some situations, and thisis one of them, the given information does not produce a unique triangle (that is one solution only)and so it is always advisable to draw a reasonably accurate diagram before proceeding further. It iseasily seen that in this example two solutions are possible.

..........................................................................................................................................................................................................

......................................

......................................

......................................

......................................

......................................

......................................

..................................................................................................................................................................

A

BFigure 5 (a)

C

20......................................

..........................................................................................................................................................................................................................................................................................................................................................................................................................

......................................

......................................

......................................

......................................

......................................

......................................

..................................................................................................................................................................

A

BFigure 5 (b)

C

20......................................

sinB =b

csinC =

2

1sin 20◦ = .6840

Hence B = 43.1602◦ (= .7533), see figure 5(b), or B = 180◦ − 43.1602◦ = 136.8398◦ (= 2.3883), asshown on figure 5(a). It follows that

A = 180◦ −B − C = 180◦ − 43.1602◦ − 20◦ = 116.8398◦, (= 2.0392)

or 180◦ − 136.8398◦ − 20◦ = 23.1602◦, (= .4042).

Finally we may find a from the cosine formula.

a =√b2 + c2 − 2bc cosA =

√22 + 12 − 2× 2× 1× cos 116.8398◦ = 2.60883 cm

or√

22 + 12 − 2× 2× 1× cos 23.1602◦ = 1.1499 cm.

5. The area of a triangle The area of a triangle is half that of a rectangle on the same base—that isto say, it is one half of the length of its base times its height.

Hence, using figure 4 and remembering that h = b sinC,

Area of 4ABC =1

2× base× height =

1

2× a× h =

1

2ab sinC .

In a similar way it can also be shown that

Area of 4ABC =1

2bc sinA =

1

2ca sinB .

Although these formulae could be remembered and used it is usually easier to proceed from first principlesby calculating the height directly (see example below).

Example 9. Find the area of the triangle ABC if b = 1 cm, c = 2 cm, C = 20◦.

This is the same triangle as in Example 8 but with the lengths of the given sides interchanged. Butthis time figure 6 confirms the answer is unique.

.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................................

......................................

..........................................................................................................................................................................................................................................................................................................................

A

B C

20

Figure 6

.......

.......

.......

.................

sinB =b

csinC =

1

2sin 20◦ = .1710

– 6 –

Hence B = 9.8466◦. It follows that

A = 180◦ −B − C = 180◦ − 9.8466◦ − 20◦ = 150.1534◦.

This enables us to find a from the cosine formula:

a =√b2 + c2 − 2bc cosA =

√12 + 22 − 2× 1× 2× cos 150.1534◦.

Using a calculator we deduce a = 2.9102 cm.

The height h is easily found fromh

1= sin 20◦, and hence the area of the triangle is given by

Area of 4ABC =1

2× a× h

=1

2× 2.9102× 1× sin 20◦

= .4977 cm2 .

***Do Exercise 3. Solve the right-angled triangle shown below in each of the following cases:

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. .......

.......

..........................

A

B Ca

bc

(i) a = 1 m, B = π/3; (ii) a = 2 m, B = 10◦; (iii) a = 3 m, b = 1 m.

***Do Exercise 4. Solve the triangle shown below in the following cases:

(i) c = 5.1 m, B = 47◦, A = 70◦; (ii) a = 12.1 m, b = 17.2 m, c = 18.4 m; (iii) b = 7 m, c = 3 m, C = 10◦.

...................................................................................................................................................................................................................................................................................................................................................................................................................................

..................................

..................................

..................................

..................................

..................................

..................................

..................................

..................................

............................................................................................................................................................B C

A

a

bc

***Do Exercise 5. Find the area of each of the triangles in Exercise 4.

6. Trigonometric ratios for special angles up to π/2 The angles 0◦, 30◦, 45◦, 60◦ and 90◦ (i.e. 0,π/6, π/4, π/3 and π/2) occur frequently in trigonometric work and it is helpful to memorise the correspondingvalues of sine, cosine and tangent for each of these angles. Alternatively, you should be able to sketch theappropriate triangle and deduce the value of the ratios (see below), or use your calculator!

(a) Angle 0

...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

a

bc

B C

A

θ

Figure 7

.......

.......

.......

..

– 7 –

In this case the triangle looks like figure 7 as θ approaches 0. Clearly b approaches 0 and as this happens a

and c become equal. However sin θ =b

c, cos θ =

a

c, tan θ =

b

a, and so we conclude that

sin 0 = 0, cos 0 = 1, tan 0 = 0.

(b) Angle π/6 or π/3 When θ = π/3 the 4ABC forms half of an equilateral triangle—see figure 8.Hence, if AB = 2 units, BC = 1 unit and Pythagoras’ theorem tells us that AC =

√22 − 12 =

√3 units.

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

2 2

1 1

√3

B C

A

π/3

π/6

Figure 8

.......

.......

................................................

..........

.....................................................

....

........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ................

................

................

................

................

................

................

................

................

................

................

...........

It follows that

cosπ

3=

1

2= .5, sin

π

3=

√3

2= .8660, tan

π

3=√

3 = 1.7321.

From the same triangle we can also read off the results for an angle of π/6:

cosπ

6=

√3

2= .8660, sin

π

6=

1

2= .5, tan

π

6=

1√3

= .5774.

(c) Angle π/4 This time the picture is as in figure 9, with a = b = 1 and c =√

2.

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

1

1√

2

B C

A

π/4

Figure 9

.......

.......

...............................................

It follows that:

cosπ

4=

1√2

= .7071, sinπ

4=

1√2

= .7071, tanπ

4= 1.

(d) Angle π/2 This case is much the same as (a), but with the triangle the other way round, as shown infigure 10. This time, however, it is the angle at A which approaches 0 so that a approaches zero length andb and c eventually become equal. Thus

cosπ

2= 0, sin

π

2= 1.

– 8 –

As θ becomes nearer and nearer to π/2 you can also see that tan(π/2) gets larger and larger. We write

tan θ →∞ as θ → π

2.

If you calculate tan(π/2), your calculator will show an error.

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

a

bc

B C

A

θ

Figure 10

.......

..........................

.........

7. Angles between π/2 and π cos θ, sin θ and tan θ have so far only been defined when θ is an acuteangle, although in using your calculator to evaluate some of your results you have used larger angles.

If θ lies between π/2 and π, figure 11 shows what we must do.

........................................................................................................................................................................................ ..................................................................................................................................................................................................................................................................................

B C

A

Fxθα

Figure 11

.......

.........................

.........................................

.......

.................................

................................................................................................................................................................................................

Choose BC to be along the positive x-axis of a coordinate system with B at the origin.

We define the three trigonometrical ratios cos θ, sin θ and tan θ to have the same numerical values respectivelyas those of cosα, sinα and tanα but, since BF is in the negative direction of the x-axis (shown in the figurealong the direction of BC), we put a minus sign in front of each of the ratios which involve this length.Notice that other distances are still treated as positive. Thus:

cos θ = −BFBA

= − cosα , sin θ =FA

BA= + sinα , tan θ = −FA

BF= − tanα .

Since α+ θ = π, it follows that α = π − θ and the above formulae can be written

cos θ = − cos(π − θ), sin θ = sin(π − θ), tan θ = − tan(π − θ).

– 9 –

8. Angles between π and 2π and > 2π or < 0

...................................................................................................................................................................... ...........................................................................................................................................................................................................................................................

B

C

A

F x

θ

α

Figure 12

................................

.......

................................................

...............................................................................................................................................................................

.................................................................................................................................................................................................................................................................................................................... .........................................................................................................................................................................................................................................................................................................................................

B C

A

F xθ α

Figure 13

..................................

.......

...............................

............................................................................................................................................................................................................

................................................................................................................................................................................

........

........

........

........

........

........

........

........

........

........

........

........

To find the trigonometric ratios for these angles, we simply extend what we did in section 7. Let us againassume that BC is along the positive x-axis of a coordinate system with B at the origin. Then the magnitudesof the trigonometric ratios for the angles ABC shown in figures 12 and 13 are the same as for the acuteangles that AB makes with this x-axis, but an appropriate sign must also be attached. As in section 7, thissign is calculated by looking at the signs attached to the lines BF and AF (note AB is always assumed tohave a positive sign). If BF is to the right, it is positive; if it is to the left, it is negative. If AF is up, it ispositive; if it is down, it is negative.

Thus in figure 12 BF is to the left and AF is down: hence cosines will be negative, sines negative andtangents (negative over negative) positive whilst, in figure 13, BF is to the right and AF down: hencecosines will be positive, sines negative and tangents negative.

Finally, if the angle is in excess of 2π we have gone full circle (at least once!) and we simply knock off asmany whole number multiples of 2π (complete revolutions) as we need in order to obtain an angle between0 and 2π and find the appropriate trigonometric ratio for this. Thus, for example,

sin9π

2= sin

(2× 2π +

π

2

)= sin

π

2= 1 .

Likewise, if the angle is less than 0, we have to add as many whole number multiples of 2π as is necessaryto put the angle into the range 0 to 2π. Thus

cos(θ + 2nπ) = cos θ, sin(θ + 2nπ) = sin θ, tan(θ + 2nπ) = tan θ,

if n is an integer. In fact tan(θ + nπ) = tan θ, but that you will see later!

Most people find it easiest to remember which signs are positive by drawing the following diagram (figure 14)in which A stands for ‘All’, S stands for ‘Sine’, T stands for ‘Tangent’ and C stands for ‘Cosine’. All othersigns are negative.

......................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

....

A

C

S

T

π/2

3π/2

0, 2π,...π

Figure 14

.......

.......

.......

.......................................................................

.......................

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................................................

– 10 –

Notice and remember the very useful results that:

cos(−θ) = cos θ, sin(−θ) = − sin θ, tan(−θ) = − tan θ.

Example 10. Evaluate sin(2π/3).

2π/3 lies in the range from π/2 to π. Figure 14 therefore shows that sin(2π/3) is positive and sosin(2π/3) = + sin(π − 2π/3) = sin(π/3). Using the results obtained with figure 8 it follows thatsin(2π/3) = sin(π/3) =

√3/2 = .8660. Check that your calculator gives the latter value for both

sin(2π/3) and sin(π/3).

Example 11. Evaluate cos 2.

2 is larger than π/2 (= 1.5708) and smaller than π (= 3.1416). Hence it follows from figure 14 thatcos 2 is the negative of the cosine of π− 2 (= 1.1416). Using your calculator verify that the values forcos 2 and − cos(π − 2) are both equal to −.4161

Example 12. Evaluate cos(190◦).

190◦(= 3.3161) is between π and 3π/2. Hence, looking at figure 14, its cosine is the negative of thecosine of 3.3161−π. Check the values given by your calculator for both cos(3.3161) and − cos(3.3161−π). You’ll find that they are both the same, namely −.9848.

Example 13. Evaluate tan 5.

5 is bigger than 3π/2 (= 4.7123) and smaller than 2π (= 6.2831). Hence, from figure 14, tan 5 =− tan(2π − 5). Again, your calculator should give the same answer for both tan 5 and − tan(2π − 5),namely −3.3805.

It may be that you are more used to using degrees than radians—in that case you may prefer to checkthe range in which 5 lies by converting it to degrees (286.4789◦) and noting that this lies between270◦ and 360◦.

Example 14. Evaluate sin(33π/4).

The angle is outside the range 0 to 2π and so we have to subtract whole number multiples of 2π tobring it into range. 33π/4 = 4× 2π + 1

4π. Hence sin(33π/4) = sin(π/4) = 1/√

2.

Check that the answer (.7071) given by your calculator agrees whichever way you do the calculation—that is to say, if your calculator allows you do this type of calculation directly!

Example 15. Evaluate tan(−π/4).

tan(−π/4) = − tan(π/4) = −1.

Example 16. Evaluate sin(−4π).

−4π is equivalent to 0 for our purposes—two negative revolutions takes us back to where we started—and so sin(−4π) = sin 0 = 0.

– 11 –

9. Graphs of trigonometric functions With a graph-plotting calculator you can very easily see whatthe graphs of cos θ, sin θ and tan θ look like for θ in the range (−2π, 2π). If you don’t have such a calculator,the previous sections give sufficient values of each function for you to be able to make a good plot. The threegraphs are:

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

...........................

.......................

θ............

.......

.................

.......

.................

.......

.................

.......

.....

.........

Figure 15: Graph of cos θ

−1

+1

−2π −3π/2 −π −π/2 π/2 π 3π/2 2π

.....................................................................................................................................................................................................................................................................................................................................

.................................................................................................................................................................................................................................................

........................ ...........................................................................................................................................................................................................................................................................................................................

...................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

...........................

.......................

θ............

.......

.................

.......

.................

.......

.................

.......

.....

.........

.........

Figure 16: Graph of sin θ

−1

+1

−2π −3π/2 −π −π/2 π/2 π 3π/2 2π.......................................................................................................................................................................

............................................................................................................................................................................................................................................................

........................................................................................................................................................................... .......

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................

................................................................................................................................

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

..............................

.......................

θ..........

......

..............

......

..............

......

..............

......

....

.........

.........

.........

.........

Figure 17: Graph of tan θ

−2π −3π/2 −π −π/2 π/2 π 3π/2 2π

1

2

−1

−2

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

....................................................................................................................................................

...........................................................................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

....................................................................................................................................................

...........................................................................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

....................................................................................................................................................

...........................................................................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

....................................................................................................................................................

...........................................................................................................................................................................................................................................................................

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

As we saw in previous sections, outside the range (0, 2π) each curve repeats itself every 2π, the extensionsjoining on smoothly to the curves shown in each case. We say that these curves ‘have period 2π’. In fact,the tangent curve repeats itself every π—has period π—a fact which means that we have to take particular

– 12 –

care if we are asked to find the angle in the range (0, 2π) which has a particular value for its tangent: thereare two possible answers! We shall discuss this in more detail later.

Notice that, if the cosine curve is displaced to the right by π/2, the sine curve is obtained whereas if the sinecurve is displaced to the right by π/2 then the negative of the cosine curve is derived. This means that

sin(π

2+ θ)

= cos θ , cos(π

2+ θ)

= − sin θ ,

which are two useful results.

Notice also that the tangent curve becomes infinite (either in the positive or negative direction) at the pointsθ = 3π/2,−π/2 , π/2 , 3π/2.

***Do Exercise 6. (i) Use your calculator to find the sine, cosine and tangent of each of the following:(a) 140◦, (b) 7, (c) 700◦.

(ii) Use your calculator to find (a) cos(−20◦), (b) sin(−250◦), (c) tan(−450◦), (d) sin(−10).

(iii) Show by direct calculation that (a) cos2 120◦ + sin2 120◦ = 1, (b) cos2 410◦ − sin2 410◦ = cos 820◦, (c)−2 sin 1.5 cos 1.5 = sin(−3).

(iv) Given that π/2 ≤ α ≤ π and sinα = 24/25, find cosα and tanα without using your calculator (checkyour answers with the calculator).

10. Other trigonometric ratios There are three other ratios that you will meet. These are the recip-rocals of (i.e. one over) the cosine, sine and tangent and are called respectively the secant, cosecant andcotangent. In the standard notation:

sec θ ≡ 1

cos θ, cosec θ ≡ 1

sin θ, cot θ ≡ 1

tan θ.

Be very careful of your notation. cos−1 θ is used to mean the angle whose cosine is θ not 1/ cos θ, andsimilarly for sin−1 θ and tan−1 θ. Thus, for example, sec θ = (cos θ)−1, not cos−1 θ. cosn θ means (cos θ)n

whenever n 6= −1 but not when n = −1. Awkward, isn’t it?

It is worth pointing out that the above difficulty with notation is the main reason why the angle whose cosineis θ is often denoted by arc cos θ, instead of cos−1 θ.

Sketch for yourselves the graphs of the functions sec θ, cosec θ and cot θ, noticing in particular that, whilst−1 ≤ cos θ ≤ 1 and −1 ≤ sin θ ≤ 1 for all values of θ, the functions sec θ and cosec θ are always either ≥ 1or ≤ −1. Also notice that, unlike tan θ, cot θ is infinite at 0 and π but finite at π/2 and 3π/2.

Example 17. Evaluate sec 3.14, cosec 3.14 and tan 3.14 to 6 d.pl.

sec 3.14 = 1/ cos 3.14 = 1/− .999999 = −1.000001.

cosec 3.14 = 1/ sin 3.14 = 1/.001593 = 627.883190.

cot 3.14 = 1/ tan 3.14 = 1/− .001593 = −627.882394.

Notice how nearly numerically equal the values of sin 3.14 and tan 3.14 (and therefore of cosec 3.14and cot 3.14) are for these values of θ so close to π.

11. Two further identities You will remember that we proved in section 3 that cos2 θ+ sin2 θ = 1 andthat tan θ = sin θ/ cos θ for all values of θ. If we divide the first equation by cos2 θ we find

cos2 θ

cos2 θ+

sin2 θ

cos2 θ=

1

cos2 θ,

– 13 –

and making use of the second equation then gives

1 + tan2 θ = sec2 θ .

In a similar way (dividing the first equation by sin2 θ instead of cos2 θ), we find

cot2 θ + 1 = cosec2 θ .

These two formulae are useful and, if possible, should be remembered, although they do appear on theFormula Sheet (and in the Data Book). The important point is to know that formulae exist connectingtan2 θ to sec2 θ (and cot2 θ to cosec2 θ).

12. Trigonometric ratios for sums and differences of angles It is often necessary to be able to ex-press cos 2θ in terms of cos θ, a formula which is a special case of that which gives cos(A + B) in terms ofthe cosines and sines of A and B alone.

From figure 18 we can easily show that

cos(A+B) = cosA cosB − sinA sinB .

In the figure, the perpendicular, AF , from A onto BC has F as its foot and the extension of BC is of lengthe. In addition, the angle ACF is of size A+B since it is an exterior angle of the triangle ABC.

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ .........................................................................................................................................................................................................................................................

..............................

........................................................................................................................................................................................................ .................................................................................................................................................................................................

................................

B C F

A

B A+B

A

x

a e

y hb

l

.......

.......

...................

.......

.......

.......

.......................................................

..........................

Figure 18

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

....

.............................................

In triangle ACF ,

cos(A+B) =e

b=l − ab

.

However, from triangle ABF ,l = (x+ y) cosB ,

where x = a cosB and y = b cosA. It follows that

cos(A+B) =(x+ y) cosB − a

b=

(a cosB + b cosA) cosB − ab

=a(cos2B − 1)

b+ cosA cosB = −a

bsin2B + cosA cosB

and, since the sine rule tells us thatsinB

b=

sinA

a,

we can replacea

bsinB by sinA and conclude that

cos(A+B) = cosA cosB − sinA sinB

– 14 –

as required.

If in this formula we replace A by (π/2)−A, we obtain a corresponding formula for sin(A−B) and if in thetwo formulae we now have we replace B by −B, we obtain two more formulae for cos(A−B) and sin(A+B).

From your point of view, the proofs are relatively unimportant. What matters is that you know the followingformulae are on the Formula Sheet (and in the Data Book) and you are able to use them.

cos(A+B) = cosA cosB − sinA sinB

sin(A+B) = sinA cosB + cosA sinB

cos(A−B) = cosA cosB + sinA sinB

sin(A−B) = sinA cosB − cosA sinB .

Notice the somewhat unexpected arrangement of the signs in the above equations.

The corresponding formulae for tan(A ± B) are less important since they can be obtained, if needed, fromthose for cos(A±B) and sin(A±B) by division. For example

tan(A+B) =sin(A+B)

cos(A+B),

and the expressions for sin(A+B) and cos(A+B) can then be used.

These formulae for sums and differences of angles are very useful in determining a number of the formulaequoted earlier. For example, using the formula for cos(A−B), we obtain

cos(π − θ) = cosπ cos θ + sinπ sin θ

= −1× cos θ + 0× sin θ = − cos θ .

13. Double angle formulae These are the formulae mentioned at the beginning of the previous note.If we put both A and B equal to θ in the formulae for cos(A+B) and sin(A+B), we obtain

cos 2θ = cos2 θ − sin2 θ , sin 2θ = 2 sin θ cos θ .

These results are highly important and must be committed to memory. The former may be written (usingsin2 θ + cos2 θ = 1) in the alternative forms:

cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 1− 2 sin2 θ .

These alternative identities are also very important and you must either learn them or be able to derivethem quickly.

Example 18. Prove that sin(C −D) + sin(C +D) = 2 sinC cosD.

sin(C −D) = sinC cosD − cosC sinD and

sin(C +D) = sinC cosD + cosC sinD .

By adding these two equations we obtain the required result.

– 15 –

14. Sum and difference formulae If in Example 18 we put C = 12 (A + B) and D = 1

2 (A − B), weobtain

sinA+ sinB = 2 sin1

2(A+B) cos

1

2(A−B) .

This is one of a set of four formulae which are useful in some integration problems in calculus, but not worthremembering—just know where to find them (on the Formula Sheet and in the Data Book) or how to deducethem if you need them! The other three are:

sinA− sinB = 2 cos1

2(A+B) sin

1

2(A−B) ,

cosA+ cosB = 2 cos1

2(A+B) cos

1

2(A−B) ,

cosA− cosB = −2 sin1

2(A+B) sin

1

2(A−B) ,

Notice the negative sign in the last of these.

Example 19. Express the following as products: (a) sin 6θ + sin 4θ, (b) cos 18θ − cos 2θ.

(a) sin 6θ + sin 4θ = 2 sin1

2(6θ + 4θ) cos

1

2(6θ − 4θ) = 2 sin 5θ cos θ .

(b) cos 18θ − cos 2θ = −2 sin1

2(18θ + 2θ) sin

1

2(18θ − 2θ) = −2 sin 10θ sin 8θ .

Example 20. Express cos 7θ cos 3θ as a sum or difference of two cosines.

Here we use the formula cosA+ cosB = 2 cos1

2(A+B) cos

1

2(A−B) .

We put 7θ = 12 (A+B) and 3θ = 1

2 (A−B) so that, after multiplying both equations by 2, we obtainA+B = 14θ and A−B = 6θ. Adding these two equations and dividing by two we see that A = 10θ.It follows that B = 4θ. Thus

cos 7θ cos 3θ =1

2(cos 10θ + cos 4θ) .

***Do Exercise 7. Prove the following identities: (i) (sin θ + cos θ)2 = 1 + sin 2θ,

(ii) tan(A+B) =tanA+ tanB

1− tanA tanB, (iii) tan 2θ =

2 tan θ

1− tan2 θ, (iv)

tan θ + cot θ

sec θ= cosec θ.

***Do Exercise 8. If sinα = 5/13, where α is acute, find tanα, cotα and cosecα (without calculatingα).

***Do Exercise 9. If tan θ = a/b, where θ is acute, find expressions for sin θ and sec2 θ in terms of aand b.

***Do Exercise 10. By writing 3θ = 2θ+ θ, and then using the identities for sums of angles and doubleangles, show that (i) sin 3θ = 3 sin θ − 4 sin3 θ, (ii) cos 3θ = 4 cos3 θ − 3 cos θ.

***Do Exercise 11. Using the formulae for sums and differences of angles show that

(i) sin(π

2+ θ)

= cos θ, (ii) cos(π

2+ θ)

= − sin θ, (iii) sin (π − θ) = sin θ, (iv) cos

(3π

2+ θ

)= sin θ.

– 16 –

***Do Exercise 12. Express as products of sines and/or cosines:

(i) sin 4θ+sin θ, (ii) sin 8θ−sin 6θ, (iii) cos 12θ+cos 10θ, (iv) cos θ−cos 2θ, (v) sin (π/3)+sin (π/4).

***Do Exercise 13. Express as sums or differences of trigonometric functions:

(i) 2 sin 6θ cos 2θ, (ii) 2 cos 5θ cos 3θ, (iii) 2 cos 4θ sin θ, (iv) 2 sin 7θ sin 5θ.

15. Simple trigonometric equations It is important to realise that, since sin θ and cos θ repeat them-selves every 2π, even a simple equation like

cos θ =1

2

is satisfied by an infinite number of values of θ. By referring back to figure 15, or using your graphicscalculator, you will see that not only is 1

3π a solution, but so are 2π + 13π, 4π + 1

3π, −2π + 13π and so on.

In fact it’s worse than this—even between θ = 0 and θ = 2π there are two different solutions, as you canagain see by looking at the graph of cos θ in figure 15 (or on your graphics calculator) or by using the ‘SATC’diagram in figure 14. Either method shows you that 2π− 1

3π = 53π is also a solution. We can therefore write

the general solution of this equation as

θ =1

3π + 2nπ , or θ =

5

3π + 2nπ ,

where n is any whole number (positive, negative or zero).

It makes things simpler, to define a unique inverse to each of the trigonometric functions and then calculateall other solutions to equations like the one above from it. Pocket calculators have the same problem. Theycannot come up with several different inverses at the same time and, instead, have to choose one and leaveyou to calculate the rest.

Calculators use the following convention, and you should do the same: cos−1 x is defined to be that solutionof cos θ = x which lies in the range 0 ≤ θ ≤ π, sin−1 x is that solution of sin θ = x which lies in the range− 1

2π ≤ θ ≤ 12π and tan−1 x is that solution of tan θ = x which lies in the range − 1

2π ≤ θ ≤ 12π. Notice the

ranges are not the same for each function: this is necessary in order to ensure that each possible value iscovered once and once only as can be seen from figures 15-17.

Example 21. Find the general solution of cos θ = −.3.

Your calculator will tell you that cos−1(−.3) is 1.8755. This is the solution of the equation which liesin the range [0, π]—in fact between 1

2π and π. The graph then shows you that 2π − 1.8755 = 4.4077is also a solution, giving you a second answer between 0 and 2π. The general solution is then

θ = 1.8755 + 2nπ , or θ = 4.4077 + 2nπ

for any integer n (positive, negative or zero).

Example 22. Find the general solution of sin θ = − 12

√2.

sin−1(− 12

√2) = − 1

4π = −.7854 since this is the appropriate value lying in the range [− 12π,

12π]. Your

calculator will give you the same answer. As you can see from the graph, the two solutions lyingbetween 0 and 2π are θ = π + 1

4π and 2π − 14π, i.e. θ = 5

4π and θ = 74π. The general solution is then

θ =5π

4+ 2nπ , or θ =

7π

4+ 2nπ

for integer n.

– 17 –

Example 23. Find the solution of tan θ = 1.6 which lies between π and 2π.

Your calculator gives you tan−1 1.6 = 1.0122. This is the value which lies in [− 12π,

12π]. The tangent

function (see the graph) repeats itself every π. The general solution of the equation is tan−1 θ =1.0122 + nπ and the solution we require, obtained by taking n = 1, is thus 4.1538.

Some trigonometric equations can readily be solved using the double angle formulae (section 13) and simplealgebra.

Example 24. Find the solutions to the equation 6 cos 2θ + 5 cos θ + 4 = 0 which lie in the range0 ≤ θ ≤ 2π.

The double angle formulae tell us that cos 2θ = 2 cos2 θ − 1 and hence the given equation can bewritten

6(2 cos2 θ − 1) + 5 cos θ + 4 = 0

or12 cos2 θ + 5 cos θ − 2 = 0 ,

which is a quadratic equation in cos θ. This equation factorises to give

(3 cos θ + 2)(4 cos θ − 1) = 0

and hence either 3 cos θ+ 2 = 0 or 4 cos θ− 1 = 0. It follows that cos θ = −2/3 or cos θ = 1/4 (resultswhich could also have been found from the quadratic equation using the formula).

According to the calculator, these solutions give

θ = cos−1(−2

3

)= 2.3005 or θ = cos−1

(1

4

)= 1.3181 ,

and from figure 14, or figure 15, it is clear that each of the above solutions leads to a correspondingsecond solution between 0 and 2π. Hence there are four solutions between 0 and 2π:

θ = 2.3005 , θ = 2π − 2.3005 = 3.9827 , θ = 1.3181 , θ = 2π − 1.3181 = 4.9651 .

16. Slightly more difficult trigonometric equations Equations of the form sinA ± sinB = 0 orcosA± cosB = 0 can be solved by using the appropriate sum or difference formula.

Example 25. Find the roots of the equation sin 7θ = sin θ for 0 ≤ θ ≤ 12π.

Here we use the formula for the difference of two sines:

sin 7θ − sin θ = 2 cos1

2(7θ + θ) sin

1

2(7θ − θ) = 2 cos 4θ sin 3θ .

Hence, equating this to zero, we see that either cos 4θ = 0 or sin 3θ = 0.

It follows from the first of these equations (look at the graphs!) that 4θ = 12π + nπ and from the

second that 3θ = nπ. Hence θ = 18π + 1

4nπ or 13nπ and, picking out the solutions which are in the

required range, we see (n = 0, 1 in the first equation) that θ = 18π or 3

8π or (n = 0, 1 in the secondequation) θ = 0 or 1

3π. Thus there are four solutions in the required range.

Finally, equations of the form a cos θ ± b sin θ = 0 can be solved by expressing the left-hand side in one ofthe forms A sin(θ±α) or A cos(θ±α) for suitably chosen A and α. This is an extremely important form forthe Engineer since it is expressed in terms of ‘phase’ (α) and ‘amplitude’ (A).

– 18 –

Example 26. Solve the equation 3 sin θ + 4 cos θ = 3 for 0 ≤ θ ≤ 2π.

First notice thatA sin(θ + α) = A cosα sin θ +A sinα cos θ .

Hence, if this is to be identically equal to the left-hand side of the given equation,

A cosα = 3 and A sinα = 4 .

Squaring and adding, we see that

A2(sin2 α+ cos2 α) = 32 + 42

so that A = 5. It follows that cosα = 35 and sinα = 4

5 , so that, by calculator, α = .9273. Noticecarefully that the signs of cosα and sinα force the solution for α to lie in a single quadrant, andhence there is only one solution to this pair of equations in [0, 2π]. If we tried to use just one ofthe equations instead of both, we would get two solutions, one wrong since it would not satisfy thesecond equation!

Our equation now simplifies to5 sin(θ + .9273) = 3

orsin(θ + .9273) = 3/5 .

Once more using our calculator and looking at the sine graph, we find that θ + .9273 = .6435 orπ − (θ + .9273) = .6435. That is to say θ = −.2838 or θ = +1.5708, with general solutions θ =−.2838 + 2nπ or θ = 1.5708 + 2nπ.

Finally, choice of n = 1 in the first equation or n = 0 in the second gives the required solutions:θ = 5.9994 or 1.5708—i.e. π/2.

***Do Exercise 14. Find the values of θ in the range 0 ≤ θ ≤ 2π which satisfy the following equations:

(i) sin θ = .87, (ii) tan θ = 1.39, (iii) cos θ = .22

***Do Exercise 15. Find all values of θ which satisfy the following equations:

(i) sin 2θ = −.81, (ii) cos 3θ = −.49, (iii) tan 12θ = .63

***Do Exercise 16. Find all values of θ in the range 0 ≤ θ ≤ 180◦ which satisfy the following equations:

(i) sin 2θ = sin 60◦, (ii) cos 3θ = cos 120◦, (iii) tan2 θ = 13 .

***Do Exercise 17. Find all solutions of the following equations which lie in the range 0 ≤ θ ≤ π:

(i) 2 sin2 θ + sin θ − 1 = 0, (ii) 16 tan2 θ − 24 tan θ + 9 = 0, (iii) sin 2θ = sin θ.

***Do Exercise 18. Find all solutions of the following equations which lie in the range 0 ≤ θ ≤ 2π:

(i) cos θ = cos 4θ, (ii) sin 3θ + sin θ = 0, (iii) 7 sin θ − 8 cos θ = 9, (iv) 5 cos θ + 12 sin θ = 4.

– 19 –

Specimen Test 2

1. Expressπ

6radians in degrees

2. In the triangle ABC find AB (correct to 4 decimal places)

...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

A B

C

8

43◦

3. If sin θ = 0.3 find (i) cosec θ, (ii) cos θ

4. In the triangle ABC find AC (correct to 4 decimal places)

........................................................................................................................................................................................................................................................................................................................................................................................

....................

....................

....................

....................

....................

....................

....................

....................

....................

.................................................................................................................................................................................................................................................................

A B

C

8

35◦ 40◦

5. In the triangle PQR find β

............................................................................................................................................................................................................................................................................................................................................................................................................

........................................

........................................

........................................

........................................

........................................

........................................

........................................

........................................

........................................

........................................

..........................................................................................................................................................................

P 10 R

Q

6

12

β

6. Sketch the graph of sinx for −2π < x < 2π

7. Express sin(π + θ) in terms of sin θ

8. Find all values of α which satisfy cosα = −0.5

9. Express cos 4θ + cos 2θ as a product of trigonometric functions

– 20 –

Worked solutions to Exercises

Note carefully that, whilst alternative answers have been given in degrees and radians, these have beencalculated independently. Owing to rounding errors, the last decimal place(s) may not agree if you tryto convert directly from an answer given in degrees to the corresponding answer in radians or vice versa.

1. (i)π

4=

(180

π

π

4

)◦=

180

4

◦= 45◦, (ii)

3π

2=

(180

π

3π

2

)◦= 270◦,

(iii)7π

3=

(180

π

7π

3

)◦= 420◦.

2. (i) 18◦ =π

18018 =

π

10= .3142, (ii) 35◦ =

π

18035 = .6109,

(iii) 350◦ =π

180350 = 6.1087, (iv) 420◦ =

π

180420 = 7.3304.

3. (i) c =a

cosB=

1

cos(π/3)=

1

.5= 2 m,

b = a tanB = tan(π/3) = 1.7321 m,

A = π −(

1

2π +B

)= π − 1

2π − 1

3π =

1

6π.

(ii) c =a

cosB=

2

cos(10◦)=

2

.9848= 2.0309 m,

b = a tanB = 2 tan(10◦) = .3527 m,

A = 180◦ − (90◦ +B) = 180◦ − 90◦ − 10◦ = 80◦.

(iii) c =√a2 + b2 =

√10 m = 3.1623 m,

tanB =b

a=

1

3. Hence, since B < 1

2π, B = tan−1 .3333 = .3218 (= 18.4349◦).

A = π −(

1

2π +B

)=

1

2π − .3218 = 1.2490, (= 71.5651◦).

4. (i) Firstly, A + B + C = 180◦. Hence, C = 180◦ − 47◦ − 70◦ = 63◦. Now we can use the sine rule to give

us a and b:a

sinA=

b

sinB=

c

sinCimplies that

a =c sinA

sinC=

5.1× .9397

.8910= 5.3787 m, b =

c sinB

sinC=

5.1× .7314

.8910= 4.1862 m.

(ii) Here we must use the cosine formula: a2 = b2 + c2 − 2bc cosA tells us that

cosA =b2 + c2 − a2

2bc=

17.22 + 18.42 − 12.12

2× 17.2× 18.4= .7710

and so A = 39.5594◦, (= .6904).

In the same way, the angle cosB is given by

cosB =c2 + a2 − b2

2ca=

18.42 + 12.12 − 17.22

2× 18.4× 12.1= .4247

and so B = 64.8655◦, (= 1.1321).

Finally, C = 180◦ −A−B = 180◦ − 39.5594◦ − 64.8655◦ = 75.5751◦, (= 1.3190).

(iii) Given two sides and non-included angle, we must use the sine formula—and the answer is not unique(draw a diagram to confirm this)!

sinB =b

csinC =

7

3sin 10◦ = .4052

– 21 –

Hence B = 23.9023◦, (= .4172) or B = 180◦ − 23.9023◦ = 156.0977◦, (= 2.7244). It follows that

A = 180◦−B−C = 180◦−23.9023◦−10◦ = 146.0977◦, (= 2.5499) or 180◦−156.0977◦−10◦ = 13.9023◦,(= .2426).

Finally we may find a from the cosine formula.

a =√b2 + c2 − 2bc cosA =

√72 + 32 − 2× 7× 3× cos 146.0977◦ =

√92.8596 = 9.6364 m

or√

72 + 32 − 2× 7× 3× cos 13.9023◦ =√

17.2303 = 4.1509 m.

5. The area in each case is given by, for example 12 × base× height = 1

2 × a× b sinC = 12ab sinC. Thus the

answers are

(i) 12 × 5.3787× 4.1862× sin 63◦ = 10.0311 m2,

(ii) 12 × 12.1× 17.2× sin 75.5751◦ = 100.7795 m2,

(iii) 12 × 9.6364× 7× sin 10◦ = 5.8567 m2 or 1

2 × 4.1509× 7× sin 10◦ = 2.5228 m2.

6. (i) (a) .6428, −.7660, −.8391, (b) .6570, .7539, .8714, (c) −.3420, .9397, −.3640.

(ii) (a) .9397, (b) .9397, (c) ∞, (d) .5440.

(iii) (a) cos2 120◦ + sin2 120◦ = .25 + .75 = 1,

(b) cos2 410◦ − sin2 410◦ = .4132− .5868 = −.1736 = cos 820◦

(c) −2 sin 1.5 cos 1.5 = −2× .9975× .0707 = −.1411 = sin(−3).

(iv) cos2 α + sin2 α = 1. Hence cosα = ±√

1− sin2 α = ±√

1− (24/25)2 = ±7/25. However, α is obtuseand so lies between 90◦ and 180◦. It follows that cosα is negative and so cosα = −7/25.

tanα =sinα

cosα= −24/7. The sign is correct for an obtuse angle.

7. (i) (sin θ + cos θ)2 = (sin2 θ + cos2 θ) + 2 sin θ cos θ = 1 + 2 sin θ cos θ = 1 + sin 2θ .

(ii)

tan(A+B) =sin(A+B)

cos(A+B)=

sinA cosB + cosA sinB

cosA cosB − sinA sinB

On dividing top and bottom of the right hand side by cosA cosB we obtain

tan(A+B) =tanA+ tanB

1− tanA tanB.

(iii) Putting A = B = θ in part (ii), the result follows immediately.

(iv)tan θ + cot θ

sec θ=

(sin θ/ cos θ) + (cos θ/ sin θ)

(1/ cos θ)

=cos θ sin θ

((sin θ/ cos θ) + (cos θ/ sin θ)

)cos θ sin θ(1/ cos θ)

=sin2 θ + cos2 θ

sin θ

=1

sin θ= cosec θ .

8. Given that sinα = 5/13, where α is acute, the lengths of the side opposite α and the hypotenuse are 5 and13 respectively. Using Pyhthagoras’ theorem the length of the remaining side is

√132 − 52 =

√144 = 12.

Hence tanα = 5/12, cotα = 1/ tanα = 12/5 and cosecα = a/ sinα = 13/5.

– 22 –

9. Since tan θ = a/b and θ is acute, a and b are respectively the side opposite to and the side by θ in aright-angled triangle whose hypotenuse is of length

√a2 + b2. It follows that sin θ = a/

√a2 + b2. In

addition 1 + tan2 θ = sec2 θ and hence sec2 θ = (a/b)2 + 1.

10. (i)sin 3θ = sin(2θ + θ)

= sin 2θ cos θ + cos 2θ sin θ, using the formula for sin(A+B)

= 2 sin θ cos2 θ + (1− 2 sin2 θ) sin θ, using the formulae for sin 2θ and cos 2θ

= 2 sin θ(1− sin2 θ) + (1− 2 sin2 θ) sin θ

= 3 sin θ − 4 sin3 θ .

(ii)cos 3θ = cos(2θ + θ)

= cos 2θ cos θ − sin 2θ sin θ, using the formula for cos(A+B)

= (2 cos2 θ − 1) cos θ − 2 sin2 θ cos θ, using the formulae for sin 2θ and cos 2θ

= (2 cos2 θ − 1) cos θ − 2(1− cos2 θ) cos θ

= 4 cos3 θ − 3 cos θ .

11. (i) sin(π

2+ θ)

= sinπ

2cos θ + cos

π

2sin θ = 1× cos θ + 0× sin θ = cos θ .

(ii) cos(π

2+ θ)

= cosπ

2cos θ − sin

π

2sin θ = 0× cos θ − 1× sin θ = − sin θ .

(iii) sin (π − θ) = sinπ cos θ − cosπ sin θ = 0× cos θ − (−1)× sin θ = sin θ .

(iv) cos

(3π

2+ θ

)= cos

3π

2cos θ − sin

3π

2sin θ = 0× cos θ − (−1)× sin θ = sin θ .

12. (i) sin 4θ + sin θ = 2 sin 52θ cos 3

2θ.

(ii) sin 8θ − sin 6θ = 2 cos 7θ sin θ.

(iii) cos 12θ + cos 10θ = 2 cos 11θ cos θ.

(iv) cos θ − cos 2θ = −2 sin 32θ sin(− 1

2θ) = 2 sin 32θ sin 1

2θ.

(v) sin 13π + sin 1

4π = 2 sin 724π cos 1

24π.

13. (i) 2 sin 6θ cos 2θ = sin 8θ + sin 4θ.

(ii) 2 cos 5θ cos 3θ = cos 8θ + cos 2θ.

(iii) 2 cos 4θ sin θ = sin 5θ − sin 3θ.

(iv) 2 sin 7θ sin 5θ = − cos 12θ + cos 2θ.

14. (i) θ = 1.0552 , 2.0864 (= π − 1.0552), (= 60.4586◦ , 119.5414◦).

(ii) θ = .9472 , 4.0887 (= π + .9472), (= 54.2678◦ , 234.2678◦).

(iii) θ = 1.3490 , 4.9342 (= 2π − 1.3490, (= 77.2910◦ , 282.7090◦).

15. (i) 2θ = −.9442 + 2nπ , 4.0858 + 2nπ.

Hence θ = −.4721 + nπ , 2.0429 + nπ, (= −27.0480◦ + 180n◦ , 117.0480◦ + 180n◦).

– 23 –

(ii) 3θ = 2.0829 + 2nπ , 4.2003◦ + 2nπ.

Hence θ = .6943 + 23nπ , 1.4001 + 2

3nπ, (= 39.7802◦ + 120n◦ , 80.2198◦ + 120n◦).

(iii) 12θ = .5622 + nπ.

Hence θ = 1.1244 + 2nπ, (= 64.4219◦ + 360n◦).

16. (i) sin 2θ − sin 60◦ = 2 cos(θ + 30◦) sin(θ − 30◦) = 0.

It follows that θ+30◦ = 90◦+180n◦ so that θ = 60◦+180n◦, or θ−30◦ = 180n◦ so that θ = 30◦+180n◦.

Hence the values of θ in the required range are

θ = 30◦ , 60◦, (= .5236 , 1.0472).

(ii) cos 3θ − cos 120◦ = −2 sin(32θ + 60◦) sin(3

2θ − 60◦).

It follows that 32θ+60◦ = 180n◦ so that θ = −40◦+120n◦, or 3

2θ−60◦ = 180n◦, so that θ = 40◦+120n◦.

Hence the values of θ in the required range are

θ = 40◦ , 160◦ , 80◦, (= .6981 , 2.7925 , 1.3963◦).

(iii) tan θ = ± 1√3.

Hence the values of θ in the required range are

θ = 30◦ , 150◦, (= 16π ,

56π).

17. (i) The equation factorises to give (2 sin θ − 1)(sin θ + 1) = 0, so that either sin θ = 12 or sin θ = −1.

Hence the only solutions in the required range are

θ = 16π ,

56π, (= 30◦ , 150◦).

(ii) The equation factorises to give (4 tan θ − 3)2 = 0, so that tan θ = 34 .

Hence there is just one solution in the range, namely θ = .6435, (= 36.8699◦).

(iii) Here, sin 2θ − sin θ = sin θ(2 cos θ − 1) = 0, and so either sin θ = 0 or cos θ = 12 .

Hence the solutions in the required range are θ = 0 , π , 13π, (= 0◦ , 180◦ , 60◦).

18. (i) cos θ − cos 4θ = −2 sin 52θ sin 3

2θ = 0.

Thus 52θ = nπ so that θ = 2

5nπ, or 32θ = nπ so that θ = 2

3nπ.

It follows that the solutions in the required range are

θ = 0 , 25π ,

45π ,

65π ,

85π ,

23π ,

43π , 2π, (= 0◦ , 72◦ , 144◦ , 216◦ , 288◦ , 120◦ , 240◦ , 360◦).

(ii) sin 3θ + sin θ = 2 sin 2θ cos θ = 0.

It follows that 2θ = nπ so that θ = 12nπ, or θ = 1

2 (2n+ 1)π.

Hence the solutions in the required range are θ = 0 , 12π , π ,

32π , 2π, (= 0 , 90◦ , 180◦ , 270◦ , 360◦).

(iii) Dividing by√

72 + 82 =√

113 and writing cosα = 7/√

113, sinα = 8/√

113, so that α = .8520, we find

sin(θ − α) =9√113

.

Hence θ − α = 1.0097 + 2nπ or θ − α = 2.1319 + 2nπ.

It follows that θ = 1.8617 + 2nπ or θ = 2.9839 + 2nπ. The values in the required range are thusθ = 1.8617 , 2.9839, (= 106.6648◦ , 170.9668◦).

(iv) Dividing by√

52 + 122 = 13, and writing sinα = 5/13, cosα = 12/13, so that α = .3948, the equationbecomes

sin(θ + α) = 4/13 .

Hence θ + α = .3128 + 2nπ or θ + α = 2.8288 + 2nπ and θ = −.0820 + 2nπ or θ = 2.4340 + 2nπ. Itfollows that the values in the required range are θ = 6.2012 , 2.4340, (= 355.3003◦ , 139.4599◦).

– 24 –

Useful results

(a) Commit the following to memory. (Some of the results can easily be worked out fromgraphs, the Formula Sheet, calculators etc. but it is often necessary to have the results atyour fingertips.)

1. To convert from degrees to radians, multiply byπ

180

To convert from radians to degrees, multiply by180

π.

2. Some useful values:

30◦ =π

6, 45◦ =

π

4, 60◦ =

π

3, 90◦ =

π

2, 180◦ = π , 360◦ = 2π .

3. In the right-angled triangle shown sine, cosine and tangent are defined as follows:

sin θ =side opposite

hypotenuse≡ b

c, cos θ =

adjacent side

hypotenuse≡ a

c, tan θ =

side opposite

adjacent side≡ b

a.

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... .......

.......

...........................

a

bc

θ .................................

4.cos2 θ + sin2 θ = 1 .

5.cos(π

2− θ)

= sin θ , sin(π

2− θ)

= cos θ .

6.cos(−θ) = cos θ , sin(−θ) = − sin θ , tan(−θ) = − tan θ .

7.

tan θ =sin θ

cos θ.

8. The area of a general triangle ABC is given by

Area of 4ABC =1

2× base× height =

1

2ab sinC

or, of course, by either of the other two similar expressions.

9. The values of sine, cosine and tangents of certain important angles between 0 and π/2 are:

θ 0 π/6 π/4 π/3 π/2

sin θ 0 1/2 1/√

2√

3/2 1

cos θ 1√

3/2 1/√

2 1/2 0

tan θ 0 1/√

3 1√

3 ∞

– 25 –

10. The sine and cosine functions have period 2π, the tangent function has period π:

cos(θ + 2nπ) = cos(θ) , sin(θ + 2nπ) = sin θ , tan(θ + nπ) = tan θ ,

if n is an integer.

11. The other three trigonometric ratios are defined as follows:

sec θ ≡ 1

cos θ, cosec θ ≡ 1

sin θ, cot θ ≡ 1

tan θ.

12.cos 2θ = 2 cos2 θ − 1 , sin 2θ = 2 sin θ cos θ .

13. The diagram to help you remember the signs of the trigonometric functions of θ for all real valuesof θ is:

.................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

A

C

S

T

π/2

3π/2

0, 2π,...π ......................................................................

.........................

..........................................................................................................................................................................................................................................................................................................................

..............................................................................................

(b) The following appear on the Formula Sheet but it is very useful to know them!

1. The ‘sine’ and ‘cosine’ formulae for a general triangle ABC are

a

sinA=

b

sinB=

c

sinC,

a2 = b2 + c2 − 2bc cosA ,

with similar results giving b2 in terms of c, a, cosB and c2 in terms of a, b, cosC.

2. The graphs of the three elementary trigonometric functions between −2π and 2π are:

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.........................

.......................

θ.........

......

............

......

............

......

............

......

...

........

Graph of cos θ

−1

+1

−2π −3π/2 −π −π/2 π/2 π 3π/2 2π

....................................................................................................................................................................................................................................................................

....................................................................................................................................................................................

..................... ..........................................................................................................................................................................................................................................................

...................................................................................................................................................................................................................

– 26 –

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.........................

.......................

θ.........

......

............

......

............

......

............

......

...

........

........

Graph of sin θ

−1

+1

−2π −3π/2 −π −π/2 π/2 π 3π/2 2π.......................................................................................................................................

..............................................................................................................................................................................................

........................................................................................................................................ ........

.....................................................................................................................................................................................................................................................................................................................................................................

................................................................................................

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

........................

.......................

θ........

......

..........

......

..........

......

..........

......

..

........

........

........

........

Graph of tan θ

−2π −3π/2 −π −π/2 π/2 π 3π/2 2π

1

2

−1

−2

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

........................................................................................................................

...................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

........................................................................................................................

...................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

........................................................................................................................

...................................................................................................................................................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

.......

........................................................................................................................

...................................................................................................................................................................................................................

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

..

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

..

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

..

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

.......

.

..

3.1 + tan2 θ = sec2 θ .

4.1 + cot2 θ = cosec2 θ .

5.cos(A+B) = cosA cosB − sinA sinB ,

sin(A+B) = sinA cosB + cosA sinB .

cos(A−B) = cosA cosB + sinA sinB ,

sin(A−B) = sinA cosB − cosA sinB .

(c) Also on the Formula Sheet are:

1.

sinA+ sinB = 2 sin1

2(A+B) cos

1

2(A−B) ,

sinA− sinB = 2 cos1

2(A+B) sin

1

2(A−B) ,

cosA+ cosB = 2 cos1

2(A+B) cos

1

2(A−B) ,

cosA− cosB = −2 sin1

2(A+B) sin

1

2(A−B) ,

Solutions to the specimen test are on the website.

– 27 –