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Examples

1) Solve for is 2 cos = 0,646cos = 0,323

Now: = cos1(0,323) + k360 = 108,8 + k360,k

= 108,8 + k360, k

2) Solve for if cos 3= 0,632. Get the angle 3= cos1( 0,632) + k360: k

3= 129,2 + k360.

We dont want 3we want , so = 43,06 + k120k. Notice here that the

period of cos 3is no longer 360, but 360_3

= 120

3) Solve for if cos( 40) = tan 22.

cos( 40) = [0,4040262258]

Leave that answer in the calculator and press shift cos to get the angle.

40= 66,2 + k360= 4066,2+ k360

= (106,2 or 26,2)+ k360, k

4) Solve for if cos ( 50) = cos 2. You have the angle. Spread out

cos ( 50) = cos 2

So: 50= 2+ k360

= 2= 50+ k360

Thus 3= 50+ k360 or = 50+ k360

= 16,7+ k120 or = 50+ k360, k

5) Solve for

if cos

2

2

=

1

_4

cos 2= 1_2

2= cos1(1_2)+ k360 or 2= cos1( 1_2)+ k360

2= 60+ k360 or 2= 120+ k360

= 30+ k180 or = 60+ k180, k

6) cos 3= cos

We need to remove the negative by reading from the left to the right.

cos 3= cos and is thus negative.

So cos 3

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+ = 450+ k90 (1) = 90+ k120 (2), k

Since [ 180; 90]

For 1: = 45; 135

From 2: = 90; 30; 150

Co-ratio equations

7. Solve for if cos ( 10) = sin 2 .

If this was cos ( 10) = cos 2 as in a similar example (4) above, then the

ratios are already balanced. So we would only focus on the angles.

Here our duty is to get those ratios the same:

Remember sine becomes cosine in vertical reduction, so cos ( 10) = cos 2

and the sin 2 has a positive sign in front of it. So we ask where is

( 10) positive?

According to the CAST rule,+

+ , it will be positive in the first and fourthquadrants.

90

2

2

So in the first quadrant sin2= cos (90 2) and in the fourth quadrant sin 2

= cos (2 90) quadrant

So: cos ( 10) = cos (90 2) or cos ( 10) = cos (2 90)

10= 90 2+ k360 or 10= 2 90+ k360

3= 100+ 360 = 80+ k360

and= (100_3 )

+ k120 = 80 + k360,k

Workbook: Lesson 16 Activity

For conclusion

This is the graph ofy= tanx(you learnt to draw it in Grade 10)

Lesson 1718

180 180

63,4

45

(45; 1)

(45; 1)

45

We will use the graph to solve tanx= 1_

2

Draw the liney= 1_2

. (Press shift tan.5, and the calculator gives us 26,6).

We need to include all the solutions. We simply add k180if kbecause the period

Lesson

Lesson

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of the tan graph is 180

General solution:x= 26,6+ k180 k.

Now lets solve tanx= 2_3

(Draw the line and see what the calculator gives you,

then + k180 k)

sox= 63,4+

k180

k

General solution for tan =p

= tan1(p) + k180k

k

Worked examples

1. Solve for

tan 3= 2,7

Press Shift tan (2,7) to get the angle and write down the general solution. 3= 69,7 + k180 Divide by 3

= 23,2 + k60, k

2. Solve for if [180; 180]

tan2= 0,81

tan = 0,9 Split them

tan = 0,9 or tan = 0,9 Find angles for the general solution

= 42+ k180 or = 42+ k180, k

Substitute for kto find angles between [180; 180]

{138; 42; 42; 138}

3. Solve for if tan 3= tan(+ 20)

tan 3= tan (+ 20)

tan 3

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2x= 33,7+ k180

x= 16,85+ k90

General solution for sin =p

sinis +ve in quadrants 1 and 2

So if sin =p

then:

= {sin1(p)

180 sin1(p)}+ k360; k

Example 1

Find the general solution for insin= 1_2

:

sin = 1_2

=

{sin1

(

1

_

2

)

180 sin1(1_2)}+ k360;

= {30150} + k360, k

Example 2

sin 2= 3_4

2=

{

sin1( 3_4)

180 sin1(

3_

4)}

+ k360

2= { 48,6180( 48,6)}+ k360

2= { 48,6228,6}+ k360

= { 24,3114,3}+ k180, kExample 3

Solve for if sin 2= 0,8 2= 53,13+ k360 or 2= 180 (53,13) + k360

= 26,56+ k180 or 2= 233,13+ k360

= 116,56+ k180, k

Co-ratio equations

Solve forxif sin 2x= cos 3x

In front of the cos, there is a + sign

So sin 2x>0 and according to the CAST rule,+

+ , this is in the first andsecond quadrants.

So we need to make the cosine a sine, and we can only do this through verticalreduction. That iscos3xbecomessin(90 3x)

Solution

Example

Example

Example

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So:sin2x= sin(90 3x)

2x= 90 3x + k360

2x 3x= 90 + k360 or 2x+ 3x = 90 + k360

x= 90 + k360 5x= 90 + k360

x= 90 + k360x= 18 + k72;k

Trigonometric equations involving factorisation.

a) Common factor problems are those that usually have two terms.

Example

Solve forx:3cosxsinx= 2cosx

Do not divide bycosxbecause you will lose solutions

3cosxsinx 2cosx= 0. Write in the form where one side is zero.

cosx(3sinx 2)= 0. Take out a common factor

cosx= 0 or sinx= 2_3

x= 90 + k360 orx= 41,8 + k360 orx= 138,2 + k360 k

b) Trinomials(3 terms)

a2 a 2 is a quadratic trinomial that we can factorise.

So isa2 ab b2

In trigonometry we sometimes need to create trinomials.

Example

1. Solve for if 2 sin2+ 5 cos + 1 = 0

Look at the nonsquared term.

We need a trinomial in terms of cos that is with only cos and no sine terms.

Usesin2= 1 cos2

2(1 cos2)+ 5cos+ 1= 0 Simplify

2 2cos2+ 5cos+ 1= 0

2cos2+ 5cos+ 3= 0 Change signs

2cos2 5cos 3= 0Factorise. (This is similar to2x2 5x3= 0

wherex= cos.)

(2cos+ 1)(cos 3)= 0

2cos= 1 or cos= 3

2cos= 1 or cos= 3

cos= 1_2

= cos1(1_2)+ k360 = 120 + k360

This is a quadratic trinomial with middle term 3sincos. The algebraicequivalent is:a2+ 2ab + b2= 0. So we havecos2,and3sincos and we now need

Example

Example

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asin2to complete the trinomial. The only way we can bring that in is by using the1,since= sin2+ cos2.

Socos2 3sincos+ sin2+ cos2= 0

2. Solveforif[180 ;360] and

cos2 3sincos+ 1= 0

cos2 3sincos+ sin2+ cos2= 0

2cos2 3sincos+ sin2= 0

(2cos sin)(cos sin)= 0 Factors

2cos= sin or cos= sin Create tan

2= sin_cos

or sin_cos

= 1

tan= 2 or tan= 1

= 63,4 + k80 or = 45 + k180,k

{116,6;135;45;63,4;225;243,4}

Equations with four terms: Group in twos

Example

Solve forxif2sin2x+ 2sinxcosx+ sinx+ cosx= 0

2sinx(sinx+ cosx)+ (sinx+ cosx)= 0

(sinx+ cosx)(2sinx+ 1)= 0

sinx= cosx or sinx= 1_2

tanx= 1x= 45 + k180 orx= 30 + k360 orx= 210 + k360, k

Your Fact File

If sin =p and 1 p 1,

then = {sin1(p)

180 sin1(p)}+ k360 ,k

If cos =p and 1 p1,

then = cos1(p) + k360 ,k

If tan =p andp,

then = tan1(p) + k180 ,k

1 = sin + cos2

If equations have 2 terms

You look for a common factor

You may have to use co-ratios

You may have to form the tan ratio if each side has a cos and sine of the same

angle

If equations have three terms, it is usually a quadratic trinomial

If there are 4 terms, you have to group them in pairs

Example

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