TRIGONOMETRIC FUNCTIONS

If a rotation from the initial side to terminal side is of a revolution, the angle is said to have a measure of one degree, written as 1°. A degree is divided into 60 minutes, and a minute is divided into 60 seconds . One sixtieth of a degree is called a minute, written as 1′, and one sixtieth of a minute is called a second, written as 1″. Thus, 1° = 60′, 1′ = 60″

Some of the angles whose measures are 360°,180°, 270°, 420°, – 30°, – 420° are shown in Fig 3.3.

3.2.2 Radian measure

 There is another unit for measurement of an angle, called the radian measure. Angle subtended at the centre by an arc of length 1 unit in a unit circle (circle of radius 1 unit) is said to have a measure of 1 radian. In the Fig 3.4(i) to (iv), OA is the initial side and OB is the terminal side. The figures show the angles whose measures are 1 radian, –1 radian, 1 radian and –1 radian.

We know that the circumference of a circle of radius 1 unit is 2π. Thus, one complete revolution of the initial side subtends an angle of 2π radian.

More generally, in a circle of radius r, an arc of length r will subtend an angle of 1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre. Since in a circle of radius r, an arc of length r subtends an angle whose measure is 1 radian, an arc of length l will subtend an angle whose measure is radian. Thus, if in a circle of radius r, an arc of length l subtends an angle θ radian at the centre, we have θ = or l = r θ.

3.2.3 Relation between radian and real numbers

Consider the unit circle with centre O. Let A be any point on the circle. Consider OA as initial side of an angle. Then the length of an arc of the circle will give the radian measure of the angle which the arc will subtend at the centre of the circle. Consider the line PAQ which is tangent to the circle at A. Let the point A represent the real number zero, AP represents positive real number and AQ represents negative real numbers (Fig 3.5). If we rope the line AP in the anticlockwise direction along the circle, and AQ in the clockwise direction, then every real number will correspond to a radian measure and conversely. Thus, radian measures and real numbers can be considered as one and the same.

3.2.4 Relation between degree and radian

 Since a circle subtends at the centre an angle whose radian measure is 2π and its degree measure is 360°, it follows that

2π radian = 360° or π radian = 180°

The above relation enables us to express a radian measure in terms of degree measure and a degree measure in terms of radian measure. Using approximate value of π as , we have

1 radian =  = 57° 16′ approximately.

Also 1° = radian = 0.01746 radian approximately.

The relation between degree measures and radian measure of some common angles are given in the following table:

Notational Convention

Since angles are measured either in degrees or in radians, we adopt the convention that whenever we write angle θ°, we mean the angle whose degree measure is θ and whenever we write angle β, we mean the angle whose radian measure is β.

Radian measure = Degree measure

Degree measure = Radian measure

Example 1 Convert 40° 20′ into radian measure.

Solution We know that 180° = π radian.

Hence 40° 20′ = 40 degree = radian = radian.

Therefore 40° 20′ =  radian.

Example 2 Convert 6 radians into degree measure.

Solution We know that π radian = 180°.

Hence 6 radians = 6 degree = degree

= 343degree = 343° + minute [as 1° = 60′]

= 343° + 38′ + minute [as 1′ = 60″]

= 343° + 38′ + 10.9″ = 343°38′ 11″ approximately.

Hence 6 radians = 343° 38′ 11″ approximately.

Example 3 Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm (use ).

Solution

Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14).

Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore, in 40 minutes, the minute hand turns through of a revolution. Therefore,   or  radian. Hence, the required distance travelled is given by

Example 5 If the arcs of the same lengths in two circles subtend angles 65°and 110° at the centre, find the ratio of their radii.

Solution Let r₁ and r₂ be the radii of the two circles. Given that

Let l be the length of each of the arc. Then l = r1θ1 = r2θ2, which gives

EXERCISE 3.1

1. Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47°30′ (iii) 240° (iv) 520°

2. Find the degree measures corresponding to the following radian measures (Use ).

(i) (ii) – 4 (iii) (iv)

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use ).

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

7. Find the angle in radian through which a pendulum swings if its length is 75 cm and th e tip describes an arc of length

(i) 10 cm (ii) 15 cm (iii) 21 cm

3.3 Trigonometric Functions

In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of sides of a right angled triangle. We will now extend the definition of trigonometric ratios to any angle in terms of radian measure and study them as trigonometric functions.

Consider a unit circle with centre at origin of the coordinate axes. Let P (a, b) be any point on the circle with angle AOP = x radian, i.e., length of arc AP = x (Fig 3.6).

We define cos x = a and sin x = b

Since ∆OMP is a right triangle, we have

OM² + MP² = OP² or a² + b² = 1

Thus, for every point on the unit circle, we have

a² + b² = 1 or cos²x + sin²x = 1

Since one complete revolution subtends an angle of 2π radian at the centre of the circle, ∠AOB = , ∠AOC = π and ∠AOD =. All angles which are integral multiples of  are called quadrantal angles. The coordinates of the points A, B, C and D are, respectively, 

(1, 0), (0, 1), (–1, 0) and (0, –1). Therefore, for quadrantal angles, we have

cos 2π = 1 sin 2π = 0

Now, if we take one complete revolution from the point P, we again come back to same point P. Thus, we also observe that if x increases (or decreases) by any integral multiple of 2π, the values of sine and cosine functions do not change. Thus, sin (2nπ + x) = sin, n ∈ Z , cos (2nπ + x) = cos, n ∈ Z

Further, sin x = 0, if = 0, ± π, ± 2π , ± 3π, ..., i.e., when x is an integral multiple of π

and cos x = 0, if   ... i.e., cos x vanishes when x is an odd multiple of . Thus

sin x = 0 implies x = nπ, where n is any integer

cos x = 0 implies x = (2n + 1) , where n is any integer.

We now define other trigonometric functions in terms of sine and cosine functions:

cosec x = , x ≠ nπ, where n is any integer.

sec x = , x ≠ (2n + 1) , where n is any integer.

tan x = , x ≠ (2n +1), where n is any integer.

cot x = , x ≠ n π, where n is any integer.

We have shown that for all real x, sin² x + cos² x = 1

It follows that

1 + tan² x = sec² x (why?)

1 + cot² x = cosec² x (why?)

In earlier classes, we have discussed the values of trigonometric ratios for 0°, 30°, 45°, 60° and 90°. The values of trigonometric functions for these angles are same as that of trigonometric ratios studied in earlier classes. Thus, we have the following table:

The values of cosec x, sec x and cot x are the reciprocal of the values of sin x, cos x and tan x, respectively.

3.3.1 Sign of trigonometric functions 

Let P (a, b) be a point on the unit circle with centre at the origin such that
∠AOP = x. If ∠AOQ = – x, then the coordinates of the point Q will be (a, –b) (Fig 3.7). Therefore cos (– x) = cos x

and sin (– x) = – sin x

Since for every point P (a, b) on the unit circle, – 1 ≤ a ≤ 1 and – 1 ≤ b ≤ 1, we have – 1 ≤ cos x ≤ 1 and –1 ≤ sin x ≤ 1 for all x. We have learnt in previous classes that in the first quadrant (0 < x < ) a and b are both positive, in the second quadrant ( < x <π) a is negative and b is positive, in the third quadrant 

(π < x < ) a and b are both negative and in the fourth quadrant ( < x < 2π) a is positive and b is negative. Therefore, sin x is positive for 0 < x < π, and negative for 

π < x < 2π. Similarly, cos x is positive for 0 < x < , negative for  < x <and also positive for< x < 2π. Likewise, we can find the signs of other trigonometric functions in different quadrants. In fact, we have the following table.

3.3.2 Domain and range of trigonometric functions

From the definition of sine and cosine functions, we observe that they are defined for all real numbers. Further, we observe that for each real number x,

– 1 ≤ sin x ≤ 1 and – 1 ≤ cos x ≤ 1

Thus, domain of y = sin x and y = cos x is the set of all real numbers and range is the interval [–1, 1], i.e., – 1 ≤ y ≤ 1.

Since cosec x = , the domain of y = cosec x is the set { x : x ∈ R and
x ≠ n π, n ∈ Z} and range is the set {y : y ∈ R, y ≥ 1 or y ≤ – 1}. Similarly, the domain of y = sec x is the set {x : x ∈ R and x ≠ (2n + 1) , n ∈ Z} and range is the set

{y : y ∈ R, y ≤ – 1or y ≥ 1}. The domain of y = tan x is the set {x : x ∈ R and
x ≠ (2n + 1) , n ∈ Z} and range is the set of all real numbers. The domain of
y = cot x is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the range is the set of all real numbers.

We further observe that in the first quadrant, as x increases from 0 to , sin x increases from 0 to 1, as x increases from  to π, sin x decreases from 1 to 0. In the third quadrant, as x increases from π to, sin x decreases from 0 to –1and finally, in the fourth quadrant, sin x increases from –1 to 0 as x increases from to 2π. Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we have the following table:

Remark In the above table, the statement tan x increases from 0 to ∞ (infinity) for 0 < x <  simply means that tan x increases as x increases for 0 < x <  and assumes arbitraily large positive values as x approaches to . Similarly, to say that cosec x decreases from –1 to – ∞ (minus infinity) in the fourth quadrant means that cosec x decreases for x ∈ (, 2π) and assumes arbitrarily large negative values as x approaches to 2π. The symbols ∞ and – ∞ simply specify certain types of behaviour of functions and variables.

We have already seen that values of sin x and cos x repeats after an interval of 2π. Hence, values of cosec x and sec x will also repeat after an interval of 2π. We shall see in the next section that tan (π + x) = tan x. Hence, values of tan x will repeat after an interval of π. Since cot x is reciprocal of tan x, its values will also repeat after an interval of π. Using this knowledge and behaviour of trigonometic functions, we can sketch the graph of these functions. The graph of these functions are given above:

Example 6 If cos x = – , x lies in the third quadrant, find the values of other five trigonometric functions.

Solution Since cos x = , we have sec x =

Now sin² x + cos² x = 1, i.e., sin² x = 1 – cos² x

or sin2 x = 1 – =

Hence sin x = ±

Since x lies in third quadrant, sin x is negative. Therefore

sin x = –

which also gives

cosec x = –

tan x = = and cot x = = .

Example 7 If cot x = – , x lies in second quadrant, find the values of other five trigonometric functions.

Solution Since cot x = – , we have tan x = –

Now sec² x = 1 + tan² x = 1 + =

Hence sec x = ±

Since x lies in second quadrant, sec x will be negative. Therefore

sec x = – ,

which also gives

Further, we have

sin x = tan x cos x = (– ) (– ) =

and cosec x = = .

Example 8 Find the value of sin.

Solution We know that values of sin x repeats after an interval of 2π. Therefore

sin  = sin (10π +) = sin   = .

Example 9 Find the value of cos (–1710°).

Solution We know that values of cos x repeats after an interval of 2π or 360°. Therefore, cos (–1710°) = cos (–1710° + 5  360°)

= cos (–1710° + 1800°) = cos 90° = 0.

EXERCISE 3.2

Find the values of other five trigonometric functions in Exercises 1 to 5.

1. cos x = – , x lies in third quadrant.

2. sin x = , x lies in second quadrant.

3. cot x = , x lies in third quadrant.

4. sec x = , x lies in fourth quadrant.

5. tan x = – , x lies in second quadrant.

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765° 7. cosec (– 1410°)

8. tan      9. sin (–)

10. cot (– )

3.4 Trigonometric Functions of Sum and Difference of Two Angles

In this Section, we shall derive expressions for trigonometric functions of the sum and difference of two numbers (angles) and related expressions. The basic results in this connection are called trigonometric identities. We have seen that

1. sin (– x) = – sin x

2. cos (– x) = cos x

We shall now prove some more results:

3. cos (x + y) = cos x cos y – sin x sin y

Consider the unit circle with centre at the origin. Let be the angle P4OP1and y be the angle P₁OP₂. Then (x + y) is the angle P₄OP₂. Also let (– y) be the angle P₄OP₃. Therefore, P₁, P₂, P₃ and P₄ will have the coordinates P₁(cos x, sin x),
P₂ [cos (x + y), sin (x + y)], P₃ [cos (– y), sin (– y)] and P₄ (1, 0) (Fig 3.14).

Consider the triangles P₁OP₃ and P₂OP₄. They are congruent (Why?). Therefore, P₁P₃ and P₂P₄ are equal. By using distance formula, we get

P₁P₃² = [cos x – cos (– y)]² + [sin x – sin(–y]²

= (cos x – cos y)² + (sin x + sin y)²

= cos² x + cos² y – 2 cos x cos y + sin² x + sin² y + 2sin x sin y

= 2 – 2 (cos x cos y – sin x sin y) (Why?)

Also, P₂P₄² = [1 – cos (x + y)] 2 + [0 – sin (x + y)]²

= 1 – 2cos (x + y) + cos2 (x + y) + sin2 (x + y)

= 2 – 2 cos (x + y)

Since P₁P₃ = P₂P₄, we have P₁P₃² = P₂P₄².

Therefore, 2 –2 (cos x cos y – sin x sin y) = 2 – 2 cos (x + y).

Hence cos (x + y) = cos x cos y – sin x sin y

4. cos (x – y) = cos x cos y + sin x sin y

Replacing y by – y in identity 3, we get

cos (x + (– y)) = cos x cos (– y) – sin x sin (– y)

or cos (x – y) = cos x cos y + sin x sin y

5. cos() = sin x

If we replace x by  and y by x in Identity (4), we get

cos () = cos  cos x + sin  sin x = sin x.

6. sin () = cos x

Using the Identity 5, we have

sin () = cos  = cos x.

7. sin (x + y) = sin x cos y + cos x sin y

We know that

= sin x cos y + cos x sin y

8. sin (x – y) = sin x cos y – cos x sin y

If we replace y by –y, in the Identity 7, we get the result.

9. By taking suitable values of x and y in the identities 3, 4, 7 and 8, we get the following results:

cos  = – sin x sin  = cos x

cos (π – x) = – cos x sin (π – x) = sin x

cos (π + x) = – cos x sin (π + x) = – sin x

cos (2π – x) = cos x sin (2π – x) = – sin x

Similar results for tan x, cot x, sec x and cosec x can be obtained from the results of sin x and cos x.

10. If none of the angles x, y and (x + y) is an odd multiple of, then

tan (x + y) =

Since none of the x, y and (x + y) is an odd multiple of , it follows that cos x,

cos y and cos (x + y) are non-zero. Now

tan (x + y) = = .

Dividing numerator and denominator by cos x cos y, we have

tan (x + y) =

=

11. tan ( x – y) =

If we replace y by – y in Identity 10, we get

tan (x – y) = tan [x + (– y)]

= =

12. If none of the angles x, y and (x + y) is a multiple of π, then

cot ( x + y) =

Since, none of the x, y and (x + y) is multiple of π, we find that sin x sin y and
sin (x + y) are non-zero. Now,

cot ( x + y) =

Dividing numerator and denominator by sin x sin y, we have

cot (x + y) =

13. cot (x – y) = if none of angles x, y and x–y is a multiple of π

If we replace y by –y in identity 12, we get the result

14. cos 2x = cos²x – sin² x = 2 cos² x – 1 = 1 – 2 sin² x =

We know that

cos (x + y) = cos x cos y – sin x sin y

Replacing y by x, we get

cos 2x = cos²x – sin² x

= cos² x – (1 – cos² x) = 2 cos2x – 1

Again, cos 2x = cos² x – sin² x

= 1 – sin² x – sin² x = 1 – 2 sin² x.

We have cos 2x = cos² x – sin ² x =

Dividing numerator and denominator by cos2 x, we get

cos 2x = ,, where n is an integer

15. sin 2x = 2 sinx cos x =  , where n is an integer

We have

sin (x + y) = sin x cos y + cos x sin y

Replacing y by x, we get sin 2x = 2 sin x cos x.

Again sin 2x =

Dividing each term by cos2 x, we get

sin 2x =

16. tan 2x = if 2, where n is an integer

We know that

tan (x + y) =

Replacing y by x , we get

17. sin 3x = 3 sin x – 4 sin3 x

We have,

sin 3x = sin (2x + x)

= sin 2x cos x + cos 2x sin x

= 2 sin x cos x cos x + (1 – 2sin² x) sin x

= 2 sin x (1 – sin² x) + sin x – 2 sin³ x

= 2 sin x – 2 sin³ x + sin x – 2 sin³ x

= 3 sin x – 4 sin3 x

18. cos 3x = 4 cos³ x – 3 cos x

We have,

cos 3x = cos (2x +x)

= cos 2x cos x – sin 2x sin x

= (2cos2 x – 1) cos x – 2sin x cos x sin x

= (2cos² x – 1) cos x – 2cos x (1 – cos² x)

= 2cos³ x – cos x – 2cos x + 2 cos³ x

= 4cos³ x – 3cos x.

19. if 3, where n is an integer

We have tan 3x = tan (2x + x)

=

20. (i) cos x + cos y =

(ii) cos x – cos y = –

(iii) sin x + sin y =

(iv) sin x – sin y =

We know that

cos (x + y) = cos x cos y – sin x sin y ... (1)

and cos (x – y) = cos x cos y + sin x sin y ... (2)

Adding and subtracting (1) and (2), we get

cos (x + y) + cos(x – y) = 2 cos x cos y ... (3)

and cos (x + y) – cos (x – y) = – 2 sin x sin y ... (4)

Further sin (x + y) = sin x cos y + cos x sin y ... (5)

and sin (x – y) = sin x cos y – cos x sin y ... (6)

Adding and subtracting (5) and (6), we get

sin (x + y) + sin (x – y) = 2 sin x cos y ... (7)

sin (x + y) – sin (x – y) = 2cos x sin y ... (8)

Let x + y = θ and x – y = φ. Therefore

Since θ and φ can take any real values, we can replace θ by x and φ by y.

Thus, we get

cos x + cos y = 2 cos; cos x – cos y = – 2 sin, sin x + sin y = 2 sin; sin x – sin y = 2 cos.

Remark As a part of identities given in 20, we can prove the following results:

21. (i) 2 cos x cos y = cos (x + y) + cos (x – y)

(ii) –2 sin x sin y = cos (x + y) – cos (x – y)

(iii) 2 sin x cos y = sin (x + y) + sin (x – y)

(iv) 2 cos x sin y = sin (x + y) – sin (x – y).

Example 10 Prove that

Solution We have

L.H.S. =

= 3 × × 2 – 4 sin× 1 = 3 – 4 sin

= 3 – 4 × = 1 = R.H.S.

Example 11 Find the value of sin 15°.

Solution We have

sin 15° = sin (45° – 30°)

= sin 45° cos 30° – cos 45° sin 30°

=.

Example 12 Find the value of tan.

Solution We have

tan = tan = tan

= =

Example 13 Prove that

.

Solution We have

L.H.S.

Dividing the numerator and denominator by cos x cos y, we get

.

Example 14 Show that

tan 3 x tan 2 x tan x = tan 3x – tan 2 x – tan x

Solution We know that 3x = 2x + x

Therefore, tan 3x = tan (2x + x)

or

or tan 3x – tan 3x tan 2x tan x = tan 2x + tan x

or tan 3x – tan 2x – tan x = tan 3x tan 2x tan x

or tan 3x tan 2x tan x = tan 3x – tan 2x – tan x.

Example 15 Prove that

Solution Using the Identity 20(i), we have

L.H.S.

= 2 cos cos x = 2 × cos x =cos x = R.H.S.

Example 16 Prove that

Solution Using the Identities 20 (i) and 20 (iv), we get

L.H.S. = = = R.H.S.

Example 17 Prove that

Solution We have

L.H.S.

= tan x = R.H.S.

EXERCISE 3.3

Prove that:

1. sin2+ cos2– tan2 2. 2sin+ cosec2

3. 4.

5. Find the value of:

(i) sin 75° (ii) tan 15°

Prove the following:

6.

7.8.

9.

10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

11.

12. sin² 6x – sin² 4x = sin 2x sin 10x 13. cos² 2x – cos² 6x = sin 4x sin 8x

14. sin2 x + 2 sin 4x + sin 6x = 4 cos² x sin 4x

15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

16. 17.

18. 19.

20. 21.

22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

23. 24. cos 4x = 1 – 8sin² x cos² x

25. cos 6x = 32 cos⁶ x – 48cos⁴ x + 18 cos² x – 1

3.5 Trigonometric Equations

Equations involving trigonometric functions of a variable are called trigonometric equations. In this Section, we shall find the solutions of such equations. We have already learnt that the values of sin x and cos x repeat after an interval of 2π and the values of tan x repeat after an interval of π. The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solutions. The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution. We shall use ‘Z’ to denote the set of integers.

The following examples will be helpful in solving trigonometric equations:

Example 18 Find the principal solutions of the equation.

Solution We know that,

Therefore, principal solutions are .

Example 19 Find the principal solutions of the equation.

Solution We know that

Therefore, principal solutions are .

We will now find the general solutions of trigonometric equations. We have already seen that:

We shall now prove the following results:

Theorem 1 For any real numbers x and y,

sin x = sin y implies x = nπ + (–1)n y, where n ∈ Z

Proof If sin x = sin y, then

sin x – sin y = 0 or 2cos = 0

which gives cos = 0 or sin = 0

Therefore = (2n + 1) or = nπ, where n ∈ Z

i.e. x = (2n + 1) π – y or x = 2nπ + y, where n∈Z

Hence x = (2n + 1)π + (–1)^{2n + 1} y or x = 2nπ +(–1)²ⁿ y, where n ∈ Z.

Combining these two results, we get

x = nπ + (–1)ⁿ y, where n ∈ Z.

Theorem 2 For any real numbers x and y, cos x = cos y, implies x= 2nπ ± y,
where n ∈ Z

Proof If cos x = cos y, then

cos x – cos y = 0 i.e., –2 sin sin = 0

Thus sin = 0 or sin = 0

Therefore = nπ or = nπ, where n ∈ Z

i.e. x = 2nπ – y or x = 2nπ + y, where n ∈ Z

Hence x = 2nπ ± y, where n ∈ Z

Theorem 3 Prove that if x and y are not odd mulitple of , then 

tan x = tan y implies x = nπ + y, where n ∈ Z

Proof If tan x = tan y, then tan x – tan y = 0

or

which gives sin (x – y) = 0 (Why?)

Therefore x – y = nπ, i.e., x = nπ + y, where n ∈ Z

Example 20 Find the solution of sin x  = –.

Solution We have 

Hence 

Example 21 Solve.

Solution We have,

Therefore, where n ∈ Z.

Example 22 Solve 

Solution We have,

Example 23 Solve sin 2x – sin 4x + sin 6x = 0.

Solution The equation can be written as

or

i.e.

Therefore sin 4x = 0 or

i.e.

Example 24 Solve 2 cos² x + 3 sin x = 0

Solution The equation can be written as

or

or

Hence sin x = or sin x = 2

But sin x = 2 is not possible (Why?)

Therefore sin x = =.

Hence, the solution is given by

EXERCISE 3.4

Find the principal and general solutions of the following equations:

1. 2. sec x = 2

3. 4. cosec x = – 2

Find the general solution for each of the following equations:

5. cos 4 x = cos x 6. cos 3x + cos x– cos 2x = 0

7. sin 2x + cos x= 0 8. sec² 2x = 1– tan 2x

9. sin x + sin 3x + sin 5x = 0

Miscellaneous Examples

Example 25 If sin x =, cos y =, where x and y both lie in second quadrant, find the value of sin (x + y).

Solution We know that

sin (x + y) = sin x cos y + cos x sin y ... (1)

Now cos² x= 1 – sin² x = 1 – =

Therefore cos x =.

Since x lies in second quadrant, cos x is negative.

Hence cos x =

Now sin²y = 1 – cos²y = 1 –

i.e. sin y =.

Since y lies in second quadrant, hence sin y is positive. Therefore, sin y =. Substituting the values of sin x, sin y, cos x and cos y in (1), we get

= .

Example 26 Prove that

.

Solution We have

L.H.S. =

=

= =

=

= = R.H.S.

Example 27 Find the value of tan .

Solution Let  . Then .

or y² + 2y – 1 = 0

Therefore y =

Since  lies in the first quadrant,  is positve. Hence

.

Example 28 If , find the value of sin, cos and tan.

Solution Since is negative.

Also.

Therefore, sin is positive and cos is negative.

Now sec² x = 1 + tan² x =

Therefore cos² x = or cos x = (Why?)

Now = 1 – cos x =.

Therefore sin² =

or sin = (Why?)

Again 2cos² = 1+ cos x =

Therefore cos² =

or cos = (Why?)

Hence tan = = – 3.

Example 29 Prove that 

Solution We have

Miscellaneous Exercise on Chapter 3

Prove that:

1.

2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

3. (cos x + cos y)2 + (sin x – sin y)2 = 4 cos2

4. (co– cos y)2 + (si– sin y)2 = 4 sin2

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

6.

7. sin 3x + sin 2x – sin x = 4sin x cos cos

Find sin, cos and tan in each of the following :

8. tan x = , x in quadrant II 9. cos x =, x in quadrant III

10. sin x =,x in quadrant II