1 identities - online · page 42 ⇒ cos 4θ = 0 ⇒ θ = (2n + 1) π/8. illustration find the...

Kaysons Education Trigonometric Functions and Identities

Day – 1

Trigonometric Ratio and Identities

Angle and its Measurement

Angle

Let a revolving line, starting from OX, revolve about O in a plane in the direction of arrow (in

anticlockwise direction) and occupy the position OP. Then it is said to trace an angle XOP. OP is

called the final (or terminal) position of OX which is the initial position. The point O is called the

vertex. More precisely an angle may be defined as a measured of the rotation of a half ray about its

origin. An angle XOP is positive if it is traced by a ray revolving in the anticlockwise direction

and negative if it is traced by a ray revolving in the clockwise direction.

Measurement of an Angle

There are three system of measurement of an angle:

(i) Sexagesimal system or English System.

(ii) Centesimal system or French System

(iii) Circular System

Sexagesimal System (degree measure)

In this system an angle is measured in degrees, minutes and seconds. One complete revolution in

360º where one degree is written as 1º. Further 1º is divided into 60 equal parts where one part is

one minute (written as 1`). Again one minute is divided into 60 equal part and one part is one

second (written as 1``). Thus 1º = 60 minutes (or 60`) and 1` = 60 seconds (or 60``). An angle of

90º is also called a right angle.

Trigonometric Functions and

Identities

Chapter

1


Centesimal System (grade measure)

In this system an angle is measured in grades, minutes and seconds.

Here 1 right angle = 100 grades, written as 100g,

1 grade = 100 minutes, written as 100`,

and 1 minute = 100 second, written as 100``.

Circular System (radian measure)

In this system an angle is measured in radius. A radius is an angle subtended at the centre of a

circle by an arc whose length is equal to its radius.

Let AB be an arc of a circle of radius r such that length of the arc AB = r. Then ∠AOB = 1 radian

(written as 1c).

Since the whole circle subtends an angle of 360º (4 right angles) at the centre and the angles at the

centre of circle are in the ratio of substending arcs,

Number of radians in an Angle subtended by an Arc of a Circle at the Centre

Let ∠AOP = θ radian be the angle subtended by an arc AP (1 unit) of a circle at the centre O. An

arc AB = r is cut off from AP. Then,

∠AOB = 1 radiant and

Hence number of radians in .

Definitions of Trigonometric Ratio or Circular Functions

Let a revolving line starting from OX trace an angle ‘θ’ in any of the four quadrants. Let M be the

foot of the perpendicular from P upon X`OX. Regarding OM and MP as directed length (taking

OP positive), the ratio of OM, MP and OP with one another are called circular functions or

trigonometric ratio of angle θ.


Let OM = x, MP = y and OP = r > 0. The circular functions are defined as:

Trigonometric ratio (or functions) may also be defined with respect to a triangle.

In a right angled triangle ABC, ∠CAB = A and ∠BCA = 90º = π/2. AC is the base, BC the altitude

and AB is the hypotenuse. We refer to the base as the adjacent side and the altitude as the opposite

side. With reference to angle A. the six trigonometric ratio are:

is called the sine of A, and written as sin A.

is called the cosine of A, and written as cos A.

is called the tangent of A, and written as tan A.


Obviously, . The reciprocals of sine, cosine and tangent are called cosecant, secant

and cotangent of A respectively. We write these as cosec A, sec A cot A respectively. Since the

hypotenuse is the greatest side in a right angle triangle , sin A and cos A can never be greater than

unity and cosec A and sec A can never be less than unity.

Hence while tan A and cot A may take any

numerical value.

Note:-

→ All the six trigonometric functions have got a very important property in common that is of

periodicity.

→ Remember that the trigonometrically ratios are real numbers and remain same as long as angle

A is real.

Signs of Trigonometric Ratios

The following table describes the signs of various trigonometric ratios:

1st quadrant 2

nd quadrant 3

rd quadrant 4

th quadrant

All positive sin θ, cosec θ positive tan θ, cot θ positive sec θ, cos θ positive

Trigonometrical Identities

An identity is a relation which is true for all values of the independent variables. There are three

fundamental identities involving trigonometrical ratios:

(i)

(ii)

(iii)

Proof:-

(i) In triangle OPM;

Dividing both sides of (1) by OP2, we get

(ii) Dividing both sides of (1) by OM2 we get:

.

(iii) Dividing both sides of (1) by MP2 we get


Trigonometric Ratio of Standard Angles

Trigonometric Ratios of 45º

In the adjacent figure OM = PM = y and

∠POM = 45⁰. Here

,

Trigonometric Ratios of 30º

Let the revolving line trace POM = 30º. Let PM = y. Let P`

be the image of P in OX so that OPP` is an equilateral triangle.

Then OP = PP` = 2y

Trigonometric Ratio of 60°

Let a revolving line, starting from OX, trace ∠XOP = 60°.

PM is perpendicular from P and OX, P` is appoint on OX such

that ΔOPP` is equilateral.

Let OP = 2y and OP` = 2y

Trigonometric Ratio of 0o

Let a revolving line, starting from OX, trace ∠XOP = 0⁰.

So that P lies on OX. Here PM = 0.

Kaysons Education Trigonometric Equations

Page 41

Day – 1

Trigonometric Equations and Inequations

Illustration

and write the values of θ in the interval

Solution

The given equation can be written as;

For the given interval, n = 0 and n = 1.

Illustration

Solve cos θ + cos 3θ + cos 3θ + cos 5θ + cos 7θ = 0.

Solution

We have,

cos θ + cos 3θ + cos 3θ + cos 5θ + cos 7θ = 0

⇒ (cos θ + cos 7θ) + (cos 3θ + cos 5θ) = 0

⇒ 2 cos 4θ . cos 3θ + 2 cos 4θ . cos θ = 0

⇒ cos 4θ (cos 3θ + cos θ) = 0

⇒ cos 4θ (2 cos 2θ + cos θ) = 0

⇒ either cos θ = 0 ⇒ θ = (2n + 1) π/2

⇒ cos 2θ = 0 ⇒ θ = (2n + 1) π/4.

Trigonometric Equations

Chapter

2


Page 42

⇒ cos 4θ = 0 ⇒ θ = (2n + 1) π/8.

Illustration

Find the general solutions of:

Solution

Here,

Illustration

Solve 3 tan2θ – 2 sin θ = 0.

Solution

∵ We work under the condition cos θ ≠ 0, in this problem any solution for θ, for which cos θ = 0

is to be rejected.

sin θ (3 sin θ – 2 cos2 θ) = 0

∴ sin θ = 0 …(i)

or 3 sin θ – 2 cos2 θ = 0 …(ii)

From (i);

sin θ = 0

Here principal angle is θ = 0.

∴ General solution is;

Again from (ii),

3 sin θ – 2 cos2 θ = 0

⇒ 3 sin θ – 2 (1 – sin2 θ) = 0


Page 43

⇒ 2 sin2 θ + 3 sin θ – 2 = 0

⇒ (sin θ + 2) (2 sin θ – 1) = 0

⇒ sin θ = – 2 or sin θ = ½

⇒ sin θ = – 2 is not possible.

Here principal angle is

∴ The general solution is;

∴ From (A) and (B);

Solutions are,

Illustration

Solve the equation;

Solution

sin 3θ + cos 2θ = 0

Taking positive sign;

We get,

Taking (– ve) sign;

We get,

From (i) and (ii), the solutions belonging to

Note: If the equation contains one of the expressions (sin x + cos x) or (sin x – cos x) and the functions sin

2x(or the product sin x cos x) then use the fact.


Page 44

sin 2x = (sin x + cos x)2 – 1 = 1 – (sin x – cos x)

2

Illustration

Solve tan θ + tan 2θ + tan 3θ = 0.

Solution

We have,

(tan θ + tan 2θ) + tan 3θ = 0

⇒ tan (θ + 2θ) (1 – tan θ tan 2θ) + tan 3θ = 0

⇒ tan 3θ [1 – tan θ tan 2θ + 1] = 0

⇒ either tan 3θ = 0 or 2 – tan θ tan 2θ = 0

Illustration

Find all the solutions of;

4 cos2 x sin x – 2 sin

2 x = 3 sin x

Solution

4 cos2 x sin x – 2 sin

2 x = 3 sin x

4 (1 – sin2 x) sinx -2 sin

2 x – 3 sin x = 0

∴ 4 sin x – 4 sin3 x – 2 sin

2 x – 3 sin x = 0

∴ General solution set

Properties and Solutions of Triangles

Sine Law

In any triangle ABC, the sides are proportional to sines of the opposite angles,


Page 45

Proof:-

Let us consider an acute angled ∆ABC.

Where AD is perpendicular to BC, as shown in figure.

Thus from (i) and (ii),

Cosine Law

Here we express cosine of an angle in terms of sides.

Proof:-

Let us consider ∆ABC to be acute angled triangle. Where AD is perpendicular to BC, as shown

in figure.

Here,

AB2 = AD

2 + BD

2

⇒ AB2 = AD

2 + (BC – CD)

2 {as BC = BD + DC}

⇒ AB2 = AD

2 + BD

2 – 2BC. CD

⇒ AB2 = AC

2 + BC

2 – 2BC.(AC cos C)

{given AB = c, BC = a, AC = b}

Similarly,

Projection Law

Kaysons Education Logarithms and Their Properties

Page 88

Day – 1

Logarithmic Function

If a > 0 ≠ 1, then logarithm of a positive number m is defined as the index x of that power of ‘a’

which equals m. i.e. log a m = x iff ax = m.

The function defined by f(x) = log a x, > 0 a ≠ 1 is called logarithmic function. Its domain is (0, ∞)

and range is R. When base I s ‘e’ then the logarithmic function is called natural logarithmic

function and when base is 10, then it is called common logarithmic function.

Graph of Logarithmic Function

(i):- y = loga x, a > 1

When 0 < x < 1

That is, we have to chose those values of y

for which 0 < ay < 1

(ii):- y = loga x, 0 < a < 1

x = ay

When 0 < x < 1

We have to choose those values of y

for which 0 < ay < 1.

Since 0 < a < 1 ⇒ y > 0

When x > 1

We have to choose those values of y for which ay > 1

Since a < 1 ⇒ y < 0.

Properties of Logarithmic Function

Let us consider

Logarithms and Their

Properties

Chapter

3


Page 89

… …. ….

… …. ….

Multiplying these n equations

or we can write it as

(i):- loga xn = n logax

(ii):-

(iii):-

(iv):-

(v):-

Changing of Base

Illustration

If ln 2.logb 625 = log1016. ln 10 then find the value of b.

Solution

ln 2. logb54 = log102

4. ln 10

Illustration

If in right angled triangle, a and b are the lengths of sides and c is the length of hypotenuse and c –

b ≠ 1, c + b ≠ 1, then show that

logc+ba + logc–ba = 2logc+ba. logc–ba.


Page 90

Solution

We know that in a right angled triangle

c2 = a

2 + b

2

c2 – b

2 = a

2 …(i)

= 2 logc+ba. log(c – b)a = R.H.S.

Illustration

The sum of the series upto n terms

Solution

loga2 + loga4 + loga8 +….. upto n terms

= loga2 + 2 loga2 + 3 loga2 +….. n loga2

Illustration

The value of ‘b’ satisfying

(a) 30 (b) 31

(c) 32 (d) 33

Solution

Illustration

For 0 < x < 1, the value of log(1 + x) + log(1 + x2) + …. to ∞ is

(a) log (1 – x) (b) – log (1 – x)

(c) log x (d) – log x


Page 91

Solution

log [(1 + x) + log(1 + x2) + …. to ∞]

= log [1 + x + x2 + x

3 +…. to ∞]

= – log (1 – x)

Illustration

The value of x satisfying

(a) 2 (b) 6

(c) 3 (d) 4

Solution

Taking log on both sides

⇒ x = 6.


Page 92

Self Efforts

1. If logax, logbx, logcx, are in A.P. where x ≠ 1, then shown that

2. Prove that logan logbn + logbn logcn + logcn logan

3. If a2 + b

2 = 7ab,

4. Solve that following equation for x,

5.

6. Prove that xlogy – logz

.ylogz – logx

.zlogx – logy

= 1.

7.

8. Solve the equation,

x + log10(1 + 2x) = x log105 + log106.

9. The value of x satisfying inequalities log0.3(x2 + 8) > log0.39x lies in

(a) 1 < x < 8 (b) 8 < x < 13

(c) x > 8 (d) None of these

10.

(a) a > 0, b > 0 (b) a < 0, b < 0

(c) a > 0, b < 0 (d) a < 0, b > 0

11. The value of x, satisfying

(a) x = 0 (b) x = 1

(c) x = 2 (d) x = 3

12. then the value of is equal to

(c) logca (d) None of these


Page 93

13.

(a) log 2 (b) log 3

(c) 0 (d) None of these

14. Values of x satisfying the equation

15. Values of x, satisfying the equation

16.

(a) logab (b) logba

(c) – logab (d) None of these

17. The values of x, satisfying the equation for are

(a) a–2

, a–1

(b) a–1/2

, a–1

(c) a–3

, a–1

(d) a–4/3

, a–1/2

18. The value of x, satisfying the equation

(a) – 3 (b) – 4

(c) – 5 (d) – 6

Answer

6. x = 2 8. x = 1 9. 1 < x < 8 10. a > 0, b > 0

11. x = 1 12. 13. log 2 14.

15. 16. – logab 17. a–4/3, a

–1/2 18. – 4

1 identities - online · page 42 ⇒ cos 4θ = 0 ⇒ θ = (2n + 1) π/8. illustration find the...

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