O Math 10 2075

c. The co-ordinate of one end of the diamemter of the circle x2 + y2 – 7x – 5y + 17 = 0 is (3, 4). Find
the co-ordinates of the other end.

6. a. The equation of two diameters of a circle passing through the point (5, 1) are x – y = 3 and 2x +
y = 21. Find the equation of the circle.

b. Find the equation of a circle whose centre is the point of intersection of 2x + y = 4 and

2y – x = 3 and passing through the point (4, 6).EXERI

c. Find the equation of a circle whose centre is the point of intersection of the lines x – y = 4 and 2x
+ 3y + 7 = 0 and passing through the point (2, 4).

d. Find the equation of a circle whose centre is the point of intersection of 2x – y = 5 and
x + y = 7 and radius 3 units.

7. a. If the length of the diameter with end points (2, 4) and (– 2, a) is 5 units, find the equation of the
circle.

b. If the perpendicular distance of the line 3x + 4y + 10 = 0 from the centre of the circle
x2 + y2 – ax – 2y = 5 is 4 units, find the value of a.


4.6 Equation of Circle on Some Special Cases Co-ordinate
Y

Equation of Circle Touching X-axis

When a circle with centre (h, k) touches X-axis, the radius of the C(h,k)

circle will be |k|. i.e. r = |k| and the point of tangency is A(h, 0). So, the X' O r=k X
equation of the cirlce becomes (x – h)2 + (y – k)2 = k2.
A(h,0)

Example 1: Y'
Find the equation of a circle with center (– 3, 6) and touching the X-axis.

Solution:

Centre of the circle is (– 3, 6).

As the circle touches the X-axis, the radius of the circle is 6.

Then the equation of the circle is

(x – h)2 + (y – k)2 = r2

or, (x + 3)2 + (y – 6)2 = 62

or, x2 + 6x + 9 + y2 – 12y + 36 = 36

∴ x2 + y2 + 6x – 12y + 9 = 0 is the required equation. Y

Equation of Circle touching Y-axis

When a circle with centre (h, k) touches Y-axis, the radius of the B(0,k) C(h,k)
circle will be |h|. i.e. r = |h| and the point of tangency is B(0, k). So, the
equation of the cirlce becomes (x – h)2 + (y – k)2 = h2 X' O X
Y'

Equation of Circle on Some Special Cases / 147

Example 2: Find the equation of circle with centre (– 3, 6) and touching Y-axis.
Solution:

Centre of the given circle is (– 3, 6). As the circle touches the Y-axis, Y
radius of the circle is |– 3| = 3
Hence, the equation of circle is C(h,k) B(0,k)
(x – h)2 + (y – k)2 = r2
or, (x + 3)2 + (y – 6)2 = 32 A(h,0) O X
or, x2 + 6x + 9 + y2 – 12y + 36 = 9 Y'
∴ x2 + y2 + 6x – 12y + 36 = 0 is required equation of circle.

Equation of Circle Touching Both Axes

When a circle with centre (h, k) touches both the axes, the radius of X'
the circle will be r = |h| = |k| and the points of tangency are A(h, 0) and
B(0, k). So, the equation of the cirlce becomes

(x – h)2 + (y – k)2 = h2 or (x – h)2 + (y – k)2 = k2

Example 3: Find the equation of a circle with centre at (4, – 4) that touches both the axes.

Solution:

Here, the centre of the given circle is at (4, – 4). As the circle touches Y A(4,0) X
both the axes, its radius is 4 units. The equation of the circle is X' O
4
(x – h)2 + (y – k)2 = r2

or, (x – 4)2 + {y – (– 4)}2 = 42 C(4,-4)
or, (x – 4)2 + (y + 4)2 = 16

or, x2 – 8x + 16 + y2 + 8y + 16 = 16 Y'
or, x2 + y2 – 8x + 8y + 16 = 0

Which is the required equation.

Example 4: Find the equation of a circle of radius 3 units, which touches both the axes and completely

lies in the third quadrant.

Solution:

Here, the radius of the circle, r = 3 units Y

Since, the circle touches both the axes and completely lies in the O
third quadrant, its centre is at (– 3, – 3).
X' 3 X
Now, the equation of the circle is 3
(x – h)2 + (y – k)2 = r2
C(-3,-3)

or, {x – (– 3)}2 + {y – (– 3)}2 = 32 Y'
or, (x + 3)2 + (y + 3)2 = 9

148 / Optional Mathematics class 10

or, x2 + 6x + 9 + y2 + 6y + 9 = 9 R(x3,y3)
\ x2 + y2 + 6x + 6y + 9 = 0
C(h, k)
This is the required equation.
P(x1,y1) Q(x2,y2)
Equation of a Circle through the Given Three Points

Let (h, k) be the centre of the given circle which passes through
P(x1, y1), Q(x2, y2) and R(x3, y3), then PC2 = QC2 = RC2
i.e. (h – x1)2 + (k – y1)2 = (h – x2)2 + (k – y2)2 = (h – x3)2 + (k – y3)2
We can solve the above equations for h and k and then find the
equation of the circle.

Example 5: Find the equation of circle passes through the points (2, – 1), (6, 1) and (6, 3).

Solution: R(6, 3)

Let C(h, k) be the centre of the circle passes through P(2, –  1), C(h, k)
Q(6, 1) and R(6, 3), then

CP2 = CQ2 = CR2 and P(2, –1) Q(6, 1) Co-ordinate
CP2 = (h – 2)2 + (k + 1)2

CQ2 = (h – 6)2 + (k – 1)2

CR2 = (h – 6)2 + (k – 3)2 Again, CP2 = CR2
As, CP2 = CQ2 (h – 2)2 + (k + 1)2 = (h – 6)2 + (k – 3)2
(h – 2)2 + (k + 1)2 = (h – 6)2 + (k – 1)2

or, h2 – 4h + 4 + k2 + 2k + 1 or, h2 – 4h + 4 + k2 + 2k + 1
= h2 – 12h + 36 + k2 – 2k + 1 = h2 – 12h + 36 + k2 – 6k + 9

or, 8h + 4k = 32 or, 8h + 8k = 40

or, 2h + k = 8 ... ... ... (i) ∴ h + k = 5 ... ... ... (ii)

Solving equations (i) and (ii), we get h = 3 and k = 2

Centre of the circle is at (3, 2) and radius of the circle, CP = (3 – 2)2 + (2 – 1)2 = 10.
Now, the equation of the circle is
(x – h)2 + (y – k)2 = r2
or, (x – 3)2 + (y – 2)2 = ( 10)2
or, x2 – 6x + 9 + y2 – 4y + 4 = 10
∴ x2 + y2 – 6x – 4y + 3 = 0 is the required equation.
Example 6: Find the equation of the circle whose centre lies on the line x – 4y = 1 and which passes

through the points (3, 7) and (5, 5).
Solution:

Let C(h, k) be the centre of the circle passes through the points A(3, 7) and B(5, 5), then

Equation of Circle on Some Special Cases / 149

As the centre C(h, k) of the circle lies on the line x – 4y = 1,

h – 4k = 1 ... ... ... (1) A(3, 7)
C(h, k)
The locus of the point equidistant from the points A and B(5, 5)
B which passes through the centre of the circle as shown in the
figure is AC2 = BC2 x – 4y = 1

or, (h – 3)2 + (k – 7)2 = (h – 5)2 + (k – 5)2

or, h2 – 6h + 9 + k2 – 14k + 49

= h2 – 10h + 25 + k2 – 10k + 25

or, – 6h – 14k + 10h + 10k = 25 + 25 – 49 – 9

or, 4h – 4k = – 8

or, h – k = – 2 ... ... ... (2)

Now substituting the value of h from (1) in (2), we have

4k + 1 – k = – 2

or, k = – 1

Now, from equation (2), we have The radius of the circle is

h – (– 1) = – 2 r = BC = (5 + 3)2 + (5 + 1)2

or, h = – 3 or, r = 82 + 62 = 100.

The centre of the circle is (– 3, – 1). ∴ r = 10

Hence, the equation of the required circle is

(x + 3)2 + (y + 1)2 = 102

or, x2 + 6x + 9 + y2 + 2y + 1 = 100

or, x2 + y2 + 6x + 2y = 100 – 10

or, x2 + y2 + 6x + 2y = 90

Example 7: Find the equation of the circle with centre at (– 4, 3) and touching the straight line 3x +
4y + 5 = 0.

Solution:

Here, the centre of the cirlce is (h, k) = (– 4, 3). C(-4,3)

As the circle touches the line 3x + 4y + 5 = 0, the perpendicular distance
from the centre (– 4, 3) to the given line is the radius of the circle.
Therefore the radius is

r= ax1 + by1 + c = 3×(– 4) + 4 × 3 + 5 = – 12 + 12 + 5 = 5 =1 3x + 4y + 5 = 0
a2 + b2 32 + 42 25 5

Now, the equation of the circle with centre at (– 4, 3) and radius 1 unit is

(x – h)2 + (y – k)2 = r2

150 / Optional Mathematics class 10

or, {x – (– 4)}2 + (y – 3)2 = 12

or, (x + 4)2 + (y – 3)2 = 1

or, x2 + 8x + 16 + y2 – 6y + 9 = 1

or, x2 + y2 + 8x – 6y + 24 = 0

This is the required equation.
Example 8: Show that the line 3x + y + 7 10 = 0 is tangent to the circle x2 + y2 – 2x + 6y – 39 = 0.
Solution:

The given equation of circle is x2 + y2 – 2x + 6y – 39 = 0

x2 + y2 – 2x + 6y – 39 = 0

or, x2 – 2x + y2 + 6y = 39 C(h, k)
or, x2 – 2x + 1 + y2 + 6y + 9 = 39 + 1 + 9

or, (x – 1)2 + (y + 3)2 = 49

or, (x – 1)2 + (y + 3)2 = 72 3x + y + 7 10 = 0 Co-ordinate
Comparing it with (x – h)2 + (y – k)2 = r2

Centre of the circle = (1, – 3) and

radius (r) = 7.

The given equation of the line is 3x + y + 7 10 = 0.

Length of the perpendicular drawn from the centre (1, – 3) on the line

3x + y + 7 10 = 0 is 3×1–3+7 10 =7
9+1

Since, the length of the perpendicular = radius of the circle = 7. The straight line 3x + y + 7 10 = 0 is
tangent to the given circle.

Example 9: Find the equation of a circle in which the ends of its diameter as the intersecting points of
the line x + y – 4 = 0 and the circle x2 + y2 – 8x – 8y + 22 = 0.

Solution:

Here, the equation of the given line is

x + y – 4 = 0 or, y = 4 – x ... ... (1)

and the equation of the given circle is

x2 + y2 – 8x – 8y + 22 = 0 ... ... (2)

From equations (1) and (2), we have
x2 + (4 – x)2 – 8x – 8(4 – x) + 22 = 0
or, x2 + 16 – 8x + x2 – 8x – 32 + 8x + 22 = 0
or, 2x2 – 8x + 6 = 0

Equation of Circle on Some Special Cases / 151

or, x2 – 4x + 3 = 0

or, x2 – 3x – x + 3 = 0

or, x(x – 3) – 1(x – 3) = 0

or, (x – 3)(x – 1) = 0

Either x – 3 = 0 ⇒ x = 3

or, x – 1 = 0 ⇒ x = 1

If x = 3 then from (1), y = 4 – 3 = 1

If x = 1 then from (1), y = 4 – 1 = 3

The intersecting point of the given line and the given circle are (3, 1) and (1, 3).

Now, the equation of circle with ends of its diameter (3, 1) and (1, 3) is

(x – x1)(x – x2) + (y – y1)(y – y2) = 0
or, (x – 3)(x – 1) + (y – 1)(y – 3) = 0

or, x2 – 4x + 3 + y2 – 4y + 3 = 0

or, x2 + y2 – 4x – 4y + 6 = 0

This is the required equation.

Example 10: Find the equations of the tangents drawn from (4, 3) to the circle x2 + y2 = 4.

Solution: Y (4, 3)

Equation of the given circle is

x2 + y2 = 4 ... ... ... (i) T1

Centre of the circle is (0, 0) and radius is 2. X' O T2 X
Let the equation of tangent through (4, 3) to the given circle be
y – 3 = m(x – 4) x2 + y2 = 4
i.e. mx – y – 4m + 3 = 0 ... ... ... (ii)
Y'

where m is the slope of the tangent. m × 0 – 0 – 4m + 3 =2
Length of perpendicular from centre (0, 0) of circle to the tangent (ii) is m2 + 1

or, – m4m2 + 3 =2
+ 1

Squaring both sides, we have

or, – m4m2 + 3 2
+ 1
=4

or, 16m2 – 24m + 9 = 4m2 + 4

or, 12m2 – 24m + 5 = 0

∴ m = 24 ± (24)2 – 4 × 12 × 5 = 6 ± 21
2 × 15 6

Hence, the equation of tangent is (y – 3) = 6 ± 21  (x – 4).
6

152 / Optional Mathematics class 10

Concyclic Points A D
B C
We can draw an infinite circles passing through one or two points.
We can draw a single cirlce passing through given three non-collinear
points. If four or more points lies on the circumference of a circle then
the points are known as concyclic points. In the given figure, A, B, C
and D are concylic points.

Example 11: Prove that the points (3, 0), (5, 4), (– 1, 2) and (3, 6) are concyclic.
Solution:

Here, let the given points are A(3, 0), B(5, 4), C(– 1, 2) and D(3, 6). Co-ordinate
We find the centre of the circle passing through any three points A, B and C (say) as in previous
example 5. Then the centre is at P(2, 3). Now, the radius of the circle is
r = AP = (3 – 2)2 + (0 – 3)2 = 12 + 32 = 10
Again, the distance between P and D is PD = (3 – 2)2 + (6 – 3)2 = 12 + 32 = 10.
Now, DP is also equal to radius of the circle. So, D lies on the circumference of the circle and hence,
the given points (3, 0), (5, 4), (– 1, 2) and (3, 6) are concyclic.

Exercise 4.6A

1. Find the equation of the circle touching X-axis in the following cases.

a. Centre at (3, 5) b. Centre at (– 5, 3)

c. Centre at (– 4, – 6) d. touching at (4, 0) with radius 3 units

2. Find the equation of the circle touching Y-axis in the following cases.

a. Centre at (4, 6) b. Centre at (– 3, 6)

c. Centre at (3, – 5) d. touching at (0, – 5) with radius 6 units

3. Find the equation of the circle in the following cases.

a. Centre at (3, 3) and touches both the axes

b. Centre at (5, – 5) and touches both the axes

4. Find the equation of circle having

a. radius 4 units, touches both the axes and lies in the second quadrant

b. radius 5 units, touches both the axes and lies in the fourth quadrant

Equation of Circle on Some Special Cases / 153

Exercise 4.6B

1. Find the centre and the equation of the circle passing through the points.

a. (3, 1), (1, – 1) and (3, – 3 ) b. (1, 2), (3, 4) and (5, 2)

c. (2, – 6), (6, 4) and (– 1, 1) d. (5, 7), (6, 6) and (2, – 2)

2. a. Find the equation of the circle which passes through the points (4, 1) and (6, 5) and having its
centre on the line 4x + y = 16.

b. Find the equation of the circle which passes through the points (– 2, 0) and (0, – 2) and having its
centre on the line 2x – 3y + 1 = 0.

c. Find the equation of the circle which passes through the points (3, – 2) and (– 2, 0) and having its
centre on the line 2x – y – 3 = 0.

3. a. Find the equation of the circle with centre at (3, 4) and touching straight line 4x – 3y + 10 = 0.

b. Find the equation of the circle with centre at (3, 3) and touching the straight line 3x + 4y – 16 = 0.

4. a. Show that the line y – x = 4 is tangent to the circle x2 + y2 = 8.

b. Show that the line 3x + y – 19 = 0 is tangent to the circle x2 + y2 – 4x – 6y + 3 = 0.

c. Show that the line 3x + 4y – 35 = 0 is tangent to the circle x2 + y2 + 4x – 8y – 5 = 0.

5. a. Find the equation of a circle in which the ends of its diameter as the intersecting points of the line
x + 2y – 5 = 0 and the circle x2 + y2 + 4x – 2y – 5 = 0.

b. A line 7x + y – 33 = 0 intersects the circle with equation x2 + y2 – 2x – 2y – 23 = 0 at P and Q.
Find the equation of a circle whose one of the diameter is PQ.

6. a. Find the equation of tangent to the circle x2 + y2 = 20 at (– 2, 4).

b. Find the equation of tangent to the circle 4x2 + 4y2 + x + 2y = 17 at (1, – 2).

c. Find the equation of tangents to the circle x2 + y2 = 4 from the point (4, 0).

7. Show that the following points are concyclic.

a. (3, 3), (6, 4), (4, 0) and (7, 1) b. (1, 0), (2, – 7), (8, 1) and (9, – 6)

8. a. Deduce the equation of the circle which passes through the mid-points of the sides of a triangle
whose vertices are (0, 0), (2, 6) and (8, 4).

b. Find the equation of the circle which passes through the vertices of the triangle whose sides are,
x – y = 0, x + y = 5 and x + 3 = 0.

9. Find the equation of a circle with centre at (3, 2) and radius 3 units

a. Prove that the points A(0, 2) and B(3, – 1) lie on the circle.

b. If P is the mid point of AB and Q is the centre of the circle, find equation of PQ.

c. Find the equation of AB and angle between PQ and AB.

d. Find the area of DQAB also.

10. a. Show that the circles x2 + y2 = 2 and x2 + y2 + 3x + 3y – 8 = 0 touch internally.

b. Show that the circles x2 + y2 – 4x + 6y + 8 = 0 and x2 + y2 – 10x – 6y + 14 = 0 touch externally.



154 / Optional Mathematics class 10

5. TRIGOsdaNfsdOfa METRY

Curriculum Contents Included Periods
35
a. Identities based on transformation
– Trigonometric ratios of compound angle
– Trigonometric ratios of multiple and submultiple angles (sin, cos and tan)
b. Transformation of trigonometric formulae
c. Trigonometric identities based on given condition
A + B + C = pc
d. Solution of trigonometric equations
e. Problems of height and distance (Including two angles of elevation and

depression)

Objectives Trigonometry

On completion of this unit, students will be able to:
find the trigonometric ratios of compound angles and also solve the problems related to these

compound angles.
find the trigonometric ratios of multiple and sub-multiple angles and also solve the related

problems.
obtain the transformation formulae and solve the related problems.
prove the condtional trigonometric identities.
solve the trigonometric equations.
solve the problems of height and distance involving two angles of elevation and depression.

Teaching Materials

From the given figure, find the value of h. A

From DABC, tan (90° – q) = h h
45
i.e. cot q = 4h5 45 B
or, tan q = h ... (i)

From DABD, tan q = 8h0 ... ... (ii)

From (i) and (ii), 45 = h q 90°– q
h 80 45m
D 80m C
or, h2 = 3600

\ h = 60m

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Trigonometry Each of 1 mark Each of 5 marks Quesitons Marks
Each of 2 marks Each of 4 marks
1 – 2 5
–1

 / 155

5 TRIGONOMETRY

5.1 Trigonometric Ratios of Multiple Angles

Trigonometrical Ratios of Multiple Angles

2A, 3A, 4A, ... ... etc. are the multiples of angle A. In this section, we will discuss the trigonometrical
ratios of angles 2A and 3A in terms of A.

Trigonometrical Ratios of an angle 2A

The trigonometrical ratios of angle 2A in terms of angle A can be derived independently by the
following geometrical proof.

Let a revolving line, starting from OX describes a circle BA P X
with centre O and radius OX. Let arc XP subtends an angle 2A at
centre i.e. ®POX = 2A. Produce XO to meet circle at B and join A
PB and draw PM⊥OX as shown. The BPX = 90° and ®PBX =
MPX = A (how ?). 2A

a. To prove sin 2A = 2 sin A cos A. OM

sin 2A = PM = 2 PM = 2 PM
OP 2 OP 2 OX

= 2 MP = 2 MP × BP
BX BP BX

= 2 sin A cos A.

b. To prove cos 2A = cos2A – sin2A

cos 2A = OM = 2 OM = 2 OM
OP 2 OP 2 OX

= 2 OM = OM + OM
BX BX

= (BM – OB) + (OX – MX) = BM – MX [ OB = OX]
BX BX

= BM – MX = BM × BP – MX × PX
BX BX BP BX PX BX

= cos A cos A – sin A sin A
∴ cos2A – sin2A.

c. To prove tan 2A = 2 tan A
1 – tan2A

tan 2A = PM = 2 PM = 2 PM
OM 2 OM OM + OM

156 / Optional Mathematics Class 10

= (BM – 2PM – MX) = 2 PM
OB) + (OX BM – MX

2 PM 2 PM
BM BM
= =
BM MX MX PM
BM – BM 1 – PM × BM

= 1 – 2 tan A A = 2 tan A
tan A tan 1 – tan2A

Trigonometrical Ratios of angle 2A in terms of A

We have already derived the formula of the compound angle (A + B). Now, we have

sin (A + B) = sin A cos B + cos A sin B

when B = A, sin (A + A) = sin A cos A + cos A sin A

∴ sin 2A = 2 sin A cos A ... ... ... (1)

Again, cos (A + B) = cos A cos B – sin A sin B

when B = A, cos (A + A) = cos A cos A – sin A sin A

∴ cos 2A = cos2A – sin2A ... ... ... (2)

cos 2A = 2 cos2A – 1 ... ... ... (3)

1 + cos 2A = 2 cos2A ... ... ... (4)

cos 2A = 1 – 2 sin2A ... ... ... (5)

1 – cos 2A = 2 sin2A ... ... ... (6) Trigonometry

and, tan (A + B) = tan A + tan B
1 – tan A tan B

when B = A, tan (A + A) = tan A + tan A
1 – tan A tan A

∴ tan 2A = 2 tan A ... ... ... (7)
1 – tan2A

cot (A + B) = cot B cot A – 1
cot B + cot A

When B = A, cot (A + A) = cot A cot A – 1
cot A + cot A

∴ cot2A – 1 ... ... ... (8)
cot 2A = 2 cot A

sin 2A and cos 2A in terms of tan A and cot A
We have, sin 2A
= 2 sin A cos A

= 2 sin A cos A = 2 sin A cos A
1 cos2A + sin2A

Trigonometric Ratios of Multiple Angles / 157

2 sin A cos A

= cos2A

cos2A + sin2A
cos2A cos2A

∴ sin 2A = 2 tan A ... ... ... (9)
1 + tan2A

Similarly, cot2A – 1 ... ... ... (10)
cot 2A = 2 cot A ... ... ... (11)

cos 2A = cos2A – sin2A = cos2A – sin2A
1

= cos2A – sin2A = cos2A – sin2A
cos2A + sin2A cos2A

cos2A + sin2A
cos2A

∴ cos 2A = 1 – tan2A
1 + tan2A

Similarly, cos 2A = cot2A – 1 ... ... ... (12)
cot2A + 1

From (4) and (6), we have

11 – cos 2A = tan2A ... ... ... (13)
+ cos 2A

and 1 + cos 2A ... ... ... (14)
1 – cos 2A = cot2A

Trigonometrical Ratios of angle 3A in terms of A

We have, sin 3A = sin (2A + A)

= sin 2A cos A + cos 2A sin A

= 2 sin A cos A cos A + (1 – 2 sin2A) sin A

= 2 sin A(1 – sin2A) + sin A – 2 sin3A

= 2 sin A – 2 sin3A + sin A – 2 sin3A

∴ sin 3A = 3 sin A – 4 sin3A ... ... ... (15)

4 sin3A = 3 sin A – sin 3A ... ... ... (16)

Again, cos 3A

= cos (2A + A)

= cos 2A cos A – sin 2A sin A

158 / Optional Mathematics Class 10

= (2 cos2A – 1) cos A – 2 sin A cos A sin A

= 2 cos3A – cos A – 2 cos A (1 – cos2A)
∴
= 2 cos3A – cos A – 2 cos A + 2 cos3A

cos 3A = 4 cos3A – 3 cos A ... ... ... (17)

4 cos3A = 3 cos A + cos 3A ... ... ... (18)
And, tan 3A
= tan (2A + A)

tan 2A + tan A 2 tan A + tan A 2 tan A + tan A – tan3A
1 – tan 2A tan A 1 – tan2A 1 – tan2A
= = =
2 tan A 1 – tan2A – 2 tan2A
1 – 1 – tan2A tan A 1 – tan2A

∴ tan 3A = 3 tan A – tan3­A ... ... ... (19)
1 – 3 tan2A

Similarly, cot 3A = cot3A – 3 cot A ... ... ... (20)
3 cot2A –1

Example 1: If cos θ = 4 , find the values of sin 2θ, cos 2θ and tan 2θ.
5

Solution:

Here, cos θ = 4 Trigonometry
5

sin θ = 1 – cos2q = 1 – 16 = 9 = 3
25 25 5

Then sin 2θ = 2 sin θ cos θ = 2 × 3 × 4 = 24
5 5 25

4 2 3 2 16 9 275
5 5 25 25
cos 2θ = cos2θ – sin2θ = – = – =

tan 2θ = sin 2q = 24/25 = 274.
cos 2q 7/25

Example 2: Prove the following identities:

a. cos q – sin q = sec 2θ – tan 2θ
cos q + sin q

b. 1 cos a + sin 2a a = cot α
– cos 2a + sin

c. cosec 10° – 3 sec 10° = 4

Solution:

a. RHS = sec 2θ – tan 2θ

= cos12q – sin 2q = 1 – sin 2q
cos 2q cos 2q

Trigonometric Ratios of Multiple Angles / 159

= sin2q + cos2q – 2 sin q cos q
cos2q – sin2q

= (cos q (cos q – sin q)2 sin q)
+ sin q)(cos q–

= ccooss q – sin q (LHS)
q + sin q

b. LHS = 1 cos a + sin 2a
– cos 2a + sin a

= 1c–os(1a–+22sisnin2aa) cos a
+ sin a

= 1 –co1s a (1 + 2 sin a) a
+ 2 sin2a + sin

= csoins a (1 + 2 sin a) = cot α (RHS)
a (2 sin a + 1)

c. LHS = cosec 10° – 3 sec 10°

= 1 – cos 3
sin 10° 10°

= cos 10° – 3 sin 10°
sin 10° cos 10°

=4 1 × 1 × cos 10° – 3 sin 10°
2 2 sin 10° cos 10°

cos 10° – 3 sin 10°

=4 2
2 sin 10° cos 10°

=4 1 cos 10° – 3 sin 10°
2 2

2 sin 10° cos 10°

=4 sin 30° cos 10° – cos 30° sin 10°
sin (2 × 10°)

=4 sin (30° – 10°) = 4 (RHS).
sin 20°

Example 3: Prove that:

a. 2 + 2 + 2 cos 4q = 2 cos θ

b. 4 (cos310° + sin320°) = 3 (cos 10° + sin 20°)

c. tan 7A 1 tan 4A – cot 7A 1 cot 4A = cot 3A
– +

Solution: = 2 + 2 + 2 cos 4q
a. LHS

= 2 + 2(1 + cos 4q)

= 2 + 2 × 2 cos22q

160 / Optional Mathematics Class 10

= 2 + 2 cos 2q = 2(1 + cos 2q)

= 2 × 2 cos2q = 2 cos θ (RHS).

b. LHS = 4 (cos310° + sin320°)

= 4 cos310° + 4 sin320°

= [3 cos 10° + cos (3 × 10°)] + [3 sin 20° – sin (3 × 20°)]

= 3 cos 10° + cos 30° + 3 sin 20° – sin 60°

= 3 (cos 10° + sin 20°) + 3 – 3
2 2

= 3 (cos 10° + sin 20°) (RHS)

c. LHS = tan 7A 1 tan 4A – cot 7A 1 cot 4A
– –
1
= 1 – 1 – cot 7A 1 cot 4A
cot 7A cot 4A –

= cot 4A 1 cot 7A – cot 7A 1 cot 4A
– –

cot 7A cot 4A

= cot 7A cot 4A + cot 4A 1 cot 7A
cot 4A – cot 7A –

cot 7A cot 4A + 1 Trigonometry
= cot 4A – cot 7A

= cot (7A – 4A) = cot 3A (RHS).

Example 4: Prove the following identities:

a. sin 3A – sin3A = cot A
cos3A – cos 3A

b. cos 5θ = 16 cos5θ – 20 cos3θ + 5 cos θ

c. cos6θ – sin6θ = cos 2θ 1 – 1 sin22q
4

Solution:

a. LHS = sin 3A – sin3A
cos3A – cos 3A

= 3 sin A – 4 sin3A + sin3A
cos3A – (4 cos3A – 3 cos A)

= 3 sin A – 3 sin3A
3 cos A – 3 cos3A

3 sin A(1 – sin2A)
= 3 cos A(1 – cos2A)

sin A cos2A
= cos A sin2A = cot A (RHS).

Trigonometric Ratios of Multiple Angles / 161

b. LHS = cos 5θ = cos (3θ + 2θ)

= cos 3θ cos 2θ – sin 3θ sin 2θ

= (4 cos3θ – 3 cos θ) (2 cos2θ – 1) – (3 sin θ – 4 sin3θ) 2 sin θ cos θ

= (8 cos5θ – 4 cos3θ – 6 cos3θ + 3 cos θ) – 2 cos θ sin2θ (3 – 4 sin2θ)

= (8 cos5θ – 10 cos3θ + 3 cos θ) – 2 cos θ (1 – cos2θ) (4 cos2θ – 1)

= (8 cos5θ – 10 cos3θ + 3 cos θ) – 2 cos θ (4 cos2θ – 1 – 4 cos4θ + cos2θ)

= (8 cos5θ – 10 cos3θ + 3 cos θ) – 2 cos θ (5 cos2θ – 4 cos4θ – 1)

= 8 cos5θ – 10 cos3θ + 3 cos θ – 10 cos3θ + 8 cos5θ + 2 cos θ

= 16 cos5θ – 20 cos3θ + 5 cos θ (RHS).

c. LHS = cos6θ – sin6θ

= (cos2θ)3 – (sin2θ)3

= (cos2θ – sin2θ) (cos4θ + cos2θ sin2θ + sin4θ)

= cos 2θ (cos4θ + sin4θ + cos2θ sin2θ)

= cos 2θ {(cos2θ + sin2θ)2 – 2 cos2θ sin2θ + cos2θ sin2q}

= cos 2θ (1 – cos2θ sin2θ)

= cos 2θ 1 – 1 × (2 cos q sin q)2
4
1
= cos 2θ 1 – 4 sin22q (RHS).

Example 5: If cos α = 1 p + 1 , show that cos 3α = 1 p3 + 1 .
2 p 2 p3
Solution:

We have cos α = 1 p + 1 , then
2 p

cos 3α = 4 cos3α – 3 cos α

= 4   12 p + 1   3 – 3 × 1 p + 1
p 2 p
 

= 4 × 1 p + 1 3 3 p + 1
8 p 2 p
 –

= 1   p + 1 3 – 3  p + 1  
2 p p
 

= 1   p3 + 3 p 1 p + 1   + 1 – 3  p + 1  
2 p p p3 p

= 1 p3 + 1
2 p3

Example 6: If tan2α = 1 + 2 tan2β, show that cos 2β = 1 + 2 cos 2α.

Solution:

Here, tan2α = 1 + 2 tan2β

or, tan2β = tan2a – 1
2

162 / Optional Mathematics Class 10

By componendo and dividendo,

or, 11 +– ttaann22bb = 2 – (tan2a – 1)
2 + (tan2a – 1)

or, cos 2β = 3 – tan2a
1 + tan2a

or, cos 2β = 2 + (1 – tan2a)
1+ tan2a

or, cos 2β = 1 + 2 + 1 – tan2a
tan2a 1 + tan2a

or, cos 2β = 2 + cos 2α
sec2a

or, cos 2β = 2 cos2α + cos 2α
or, cos 2β = 1 + cos 2α + cos 2α
∴ cos 2β = 1 + 2 cos 2α.

Let us Memorize

Trigonometric Ratios of Multiple Angles

sin 2A = 2 sin A cos A cos 2A = cos2A – sin2A

1. 2 tan A = 2 cot A 2. = 2 cos2A – 1 = 1 – 2 sin2A
= 1 + tan2A 1 + cot2A 1 – tan2A cot2A – 1
= 1 + tan2A = cot2A + 1
Trigonometry
3. 2 tan A 4. cot2A – 1
tan 2A = 1 – tan2A cot 2A = 2 cot A

5. sin 3A = 3 sin A – 4 sin3A 6. cos 3A = 4 cos3A – 3 cos A

7. tan 3A = 3 tan A – tan3­A 8. cot 3A = cot3A – 3 cot A
1– 3 tan2A 3 cot2A –1

9. 2 cos2A = 1 + cos 2A 10. 2 sin2A = 1 – cos 2A

11. 1 – cos 2A 12. 1 + cos 2A
tan2A = 1 + cos 2A cot2A = 1 – cos 2A

13. 4 sin3A = 3 sin A – sin 3A 14. 4 cos3A = cos 3A + 3 cos A

Exercise 5.1A

1. Find the values of sin 2A, cos 2A and tan 2A when,

a. cos A = 4 b. cos A = 153 c. sin A = 4
5 5

Trigonometric Ratios of Multiple Angles / 163

d. sin A = 175 e. tan A = 71 f. cot A = 4
3

2. Find the values of sin 3θ, cos 3θ and tan 3θ, when,

a. sin θ = 4 b. cos θ = 21 c. tan θ = 3
5 4

3. Prove the following:

a. 2 sin2(45° – A) = 1 – sin 2A b. 2 sin2(90° – A) = 1 + cos 2A

c. 2 cos2(45° – A) = 1 + sin 2A d. sin2(45° – θ) + sin2(45° + θ) = 1EXERC

e. 11 +– tan2(45° – q) = sin 2q f. 12+tatann(24(455°°––AA) ) = cos 2A
tan2(45° – q)

1 – tan2 pc – q = sin q2 h. ccoott22 pp44cc –– q2q2 –1
1 + tan2 4 4 = sin q
g.
pc q +1
4 – 4

i. tan (45° – θ) + cot (45° + θ) = 2 cos 2q
1 + sin 2q

j. tan (α + 45°) – tan (α – 45°) = 2
cos 2a

4. Prove the following identities:

a. 1 +sinco2sa2a = cot α b. 1 c–ossin2q2q = 11 +– tan q
tan q

c. 1 c+ossin2a2a = 1 – tan aa d. 11 +– ssiinn 22AA = 1 – tan A 2
1 + tan 1 + tan A

e. 1 +sicnobs b + cos 2b = cot β f. 1 + sin 2q + cos 2q = cot θ
+ sin 2b 1 + sin 2q – cos 2q

g. 11 + sin 2b – cos 2b = tan b h. 2 sin 2a – sin 4a = tan2a
+ sin 2b + cos 2b 2 sin 2a + sin 4a

i. tan θ (1 + cos 2θ) = sin 2θ j. cot α (1 – cos 2α) = sin 2α

k. (2 cos α + 1) (2 cos α – 1) = 2 cos 2α + 1

. 2 sin pc – q sin 3pc – q = cos 2θ
4 4

5. Prove the following:

a. cot A – tan A = 2 cot 2A b. 1 + tan 4A tan 2A = sec 4A

c. ccoott f + tan f = sec 2φ d. tan 2φ + sin 2φ = 4 tan f
f – tan f 1 – tan4f

e. ssiinn3qq + cos3q = 1 – 1 sin 2θ f. ssiinn3AA –– cos3A = 1 + 1 sin 2A
+ cos q 2 cos A 2

164 / Optional Mathematics Class 10

g. tatann2AA – 1 = sec 2A h. 1 – tan2A = 2 cot 2A
tan A

i. ssiinn5qq – cos 5q = 4 cos 2θ j. ssiinn 26AA – cos 6A = 2
cos q cos 2A

k. tan 1 tan q – cot 1 cot q = cot 4θ
3q + 3q +

l. tan 1 tan A – cot 1 cot A = cot 2A
3A – 3A –

6. a. If cos θ = 352, then show that cos 2θ = 11
25

b. If sin θ = 3 –1 , then prove that cos 2θ = 23.
2 2

c. If cos θ = 12, then show that cos 3θ = – 1.

d. If sin α = 23, then prove that sin 3α = 0.

Exercise 5.1B

1. a. If tan A = 1 , find the value of sin 3A. Trigonometry
3

b. If cos θ = 1 , find the value of tan 3θ.
2

2. Prove the following:

a. 11 – sin 2q = 1 c+ossin2q2q b. 11 +– ssiinn qq = tan pc + q
+ sin 2q 4 2

c. cos A – 1 + sin 2A = tan A d. sin 2A – 1 + sin 4A = cot 2A
sin A – 1 + sin 2A cos 2A – 1 + sin 4A

e. sseecc 4q – 1 = ttaann4qq f. sseecc 84AA –– 11 = ttaann 8A
2q – 1 2A

g. sin110° – cos 3 = 4 h. 3 cosec 20° – sec 20° = 4
10°

i. tan 4θ = 14–t6antaqn2–q4+tatann3q4 q j. cot 4A = cot4A – 6 cot2A – 1
4 cot3A – 4 cot A

3. Prove the following:

a. (2 cos θ + 1) (2 cos θ – 1) (2 cos 2θ – 1) = 2 cos 4θ + 1

b. tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = cot θ

c. (2 cos θ – 1) (2 cos 2θ – 1) (2 cos 4θ – 1) (2 cos 8q – 1) = 2 cos 16q + 1
2 cos q + 1

d. 2 cos θ = 2 + 2 + 2 + 2 cos 8q

e. 2 cos φ = 2 + 2 + 2 + 2 + 2 cos 16f

f. cosec 4A + cosec 8A = cot 2A – cot 8A
g. cosec 8A + cot 16A = cot 4A – cosec 16A

Trigonometric Ratios of Multiple Angles / 165

h. tan A + 2 tan 2A + 4 cot 4A = cot A

i. 2cossin3qq + 2 sin 3q + 2 sin 9q = tan 27 q – tan q
cos 9q cos 27q

4. Prove the following identities:

a. 4 (cos320º + sin310º) = 3 (cos 20º + sin 10º)

b. 4 (cos320º + sin350º) = 3 (cos 20º + sin 50º)

c. cos3α cos 3α + sin3α sin 3α = cos32α.

d. 4 sin3α cos 3α + 4 cos3α sin 3α = 3 sin 4α

e. cos2α + sin2α cos 2β = cos2β + sin2β cos 2α

f. cos 4θ = 8 cos4θ – 8 cos2θ + 1

g. sin 4A = 4 sin A cos3A – 4 cos A sin3A

h. sin 5θ = 16 sin5θ – 20 sin3θ + 5 sin θ

i. sin 8θ = 8 sin θ cos θ cos 2θ cos 4θ

j. cos8θ + sin8θ = 1 – sin22θ + 1 sin42θ
8
k. 4 (cos6α + sin6α) = 1 + 3 cos22α

. 4 (cos6θ – sin6θ) = cos32θ + 3 cos 2θ

m. (1 + sin 2θ + cos 2θ)2 = 4 cos2θ (1 + sin 2θ)

n. cos3q – cos 3q + sin3q + sin 3q = 3
cos q sin q

o. sin4θ = 3 – 1 cos 2θ + 1 cos 4θ
8 2 8

5. Prove the following trigonometric identities:

a. 4 sin θ sin q + pc sin q + 2pc = sin 3θ
3 3

b. cot θ cos (30º – θ) – sin (30º – θ) = 3 cosec θ
2

c. tan A tan (60° + A) tan (60° – A) = tan 3A

d. sin A sin (60° + A) sin (60° – A) = 1 sin 3A
4

e. tan A + tan (60° + A) + tan (120° + A) = 3 tan 3A

f. (cos 2A + cos 2B)2 + (sin 2A + sin 2B)2 = 4 cos2(A – B)

g. (cos 2A – cos 2B)2 + (sin 2A – sin 2B)2 = 4 sin2(A – B)

6. a. If tan θ = 1 and tan φ = 13, prove that cos 2θ = sin 4φ.
7
b. If tan α = ba, prove that a cos 2α + b sin 2α = a.
3
7. If tan α = 2 tan β, prove that:

a. tan (α – β) = 5 –sinco2sb2b b. tan (α + β) = 5 5 sin 2b 1
cos 2b –

166 / Optional Mathematics Class 10

8. If cos A = 1 a + 1 , prove that
2 a

a. cos 2A = 1 a2 + 1 b. cos 4A = 1 a4 + 1
2 a2 2 a4

9. If sin q = 1 m + 1 , prove that
2 m

a. cos 2q = – 1 m2 + 1 b. cos 3q = – 1 m3 + 1
2 m2 2 m3



5.2 Trigonometrical Ratios of Sub Multiple Angles

A2, A3, A4, ... ... etc. are the sub-multiples of A. Here, we will discuss about the trigonometrical ratios
of an angle A in terms of A2, A3, A4, ... ... etc. and vice versa.
A
Trigonometrical Ratios of an angle A in terms of 2

We have sin (A + B) = sin A cos B + cos A sin B.

If A = B = q2, sin q + q = sin q cos q + cos q sin q
2 2 2 2 2 2

∴ sin q = 2 sin q cos q ... ... ... (1)
2 2
Trigonometry
Again, cos (A + B) = cos A cos B – sin A sin B

If A = B = q2, cos q + q = cos q cos q – sin q sin q
2 2 2 2 2 2

∴ cos q = cos2 q – sin2 q ... ... ... (2)
2 2

cos θ = cos2 q – 1 – cos2 q
2 2

∴ cos q = 2 cos2 q – 1 ... ... ... (3)
2
... ... ... (4)
or, 1 + cos θ = 2 cos2 q2 ... ... ... (5)
q 1 + cos q
or, cos 2 = ± 2

Again, cos θ = cos2 q – sin2 q = 1 – sin2 q – sin2 q
2 2 2 2

∴ cos q = 1 – 2 sin2 q ... ... ... (6)
2 ... ... ... (7)

or, 1 – cos θ = 2 sin2 q
2

or, sin q = ± 1 – cos q ... ... ... (8)
2 2

Trigonometrical Ratios of Sub Multiple Angles / 167

From (4) and (7), we have

11 – cos q = tan2 q ... ... ... (9)
+ cos q 2 ... ... ... (10)

11 + cos q = cot2 q ... ... ... (11)
– cos q 2 ... ... ... (12)

tan (A + B) = tan A + tan B
1 – tan A tan B

If A = B = q2, tan q q tan q + tan q
2 2 2 2
+ =
1 – tan q tan q
2 2

2 tan q
2
tan q=
q
1 – tan2 2

cot2 q – 1
2
Similarly, cot q =
q
2 cot 2

sin θ and cos θ in terms of tan q and cot q
2 2

From the previous section of multiple angles, we can easily establish the formulae of sub-multiple

angles. q
2
Let 2A = θ, then A =

So, sin 2A = 12+tatannA2A 2 tan q
2
⇒ sin q =
q
1 + tan2 2

2 cot A 2 cot q
1 + cot2A 2
And, sin 2A = ⇒ sin q =
q
1 + cot2 2

Again, cos 2A = 1 – tan2A ⇒ cos q = 1 – tan2 q
1 + tan2A 1 + tan2
2
q

2

cot2A – 1 cot2 q – 1
cot2A + 1 2
and cos 2A = ⇒ cos q =
q
cot2 2 + 1

168 / Optional Mathematics Class 10

Trigonometrical ratios of angle θ in terms of q
3

The trigonometric ratios of angle θ in terms of q can be established easily from the previous section
3
of multiple angles as follows:

Let 3A = θ, then A = q3. ⇒ qq
So, sin 3A = 3 sin A – 4 sin3A sin q = 3 sin 3 – 4 sin3 3

and 4 sin3A = 3 sin A – sin 3A ⇒ 4 sin3 q = 3 sin q – sin q
3 3

Again, cos 3A = 4 cos3A – 3 cos A ⇒ qq
cos q = 4 cos3 3 – 3 cos 3

and 4 cos3A = cos 3A + 3 cos A ⇒ 4 cos3 q = cos q + 3 cos q
3 3

3 tan A – tan3A 3 tan q – tan3 q
1– 3 tan2A 3 3
As, tan 3A = ⇒ tan q =
q
1 – 3 tan2 3

cot3A – 3 cot A cot3 q – 3 cot q Trigonometry
3 cot2A –1 3 3
and cot 3A = ⇒ cot q =
q
3 cot2 3 – 1

Example 1: If sin 30° = 1 , find the values of sin 15° and tan 15°.
2

Solution:

Here, sin 30° = 12, 1 – 1 = 3
cos 30° = cos230° = 1 – sin230° = 4 2

Now, 2 sin2 30° = 1 – cos 30° = 1 – 3
2 2

or, 2 sin215° = 2 – 3
2

or, sin215° = 2 – 3 = 4 –2 3 = 3 – 23 + 1
4 8 8

or, sin215° = 3 2 – 2. 3.1 + 12 = 3 – 12
2 22 2 22

∴ sin 15° = 3 –1 [As 15° lies in first quadrant, sin 15° is positive]
2 2

Again, tan2 q = 1 – cos q
2 1 + cos q

Trigonometrical Ratios of Sub Multiple Angles / 169

or, tan2 30° = 1 – cos 30° = 1– 3 = 2– 3
2 1 + cos 30° 1+ 2 2+ 3

3
2
2 – 32
or, tan215° = 2– 3 × 2– 3 = 4–3
2+ 3 2– 3

∴ tan 15° = 2 – 3 [As 15° lies in first quadrant, tan 15° is positive]

Example 2: Show that:

sin3 q + cos3 q
sin 2 2
a. = 1 – sin θ
q q
2 + cos 2

b. cos2 pc + q – sin2 pc + q = sin q
4 4 4 4 2

Solution:

sin3 q + cos3 q sin q + cos q sin2 q – sin q cos q + cos2 q
2 2 2 2 2 2 2 2
a. LHS = =
q q sin q + cos q
sin 2 + cos 2 2 2

= sin2 q + cos2 q – 1 × 2 sin q cos q =1 – 1 sin θ (RHS)
2 2 2 2 2 2

b. LHS = cos2 pc – q – sin2 pc – q
4 4 4 4

= cos 2 pc – q = cos pc – q = sin q (RHS)
4 4 2 2 2

Example 3: Show that:
a. q
Solution: cos 2 – 1 + sin q = tan q2 b. cos4 pc + cos4 3pc + cos4 5pc + cos4 7pc = 3
1 + sin q 8 8 8 8 2
sin q –
2

cos q – 1 + sin q cos q – c o s 2 q + s in2 q + 2 cos q sin q
2 1 + sin q 2 2 2 2 2
c. LHS = =
q q q q q q
sin 2 – sin 2 – co s 2 2 + si n2 2 + 2 cos 2 sin 2

cos q – cos q + sin q 2 cos q – cos q + sin q cos q – cos q – sin q
2 2 2 2 2 2 2 2 2
= = =
q q q 2 sin q – q q sin q – cos q – sin q
sin 2 – cos 2 + sin 2 2 cos 2 + sin 2 2 2 2

– sin q q
2 2
= = tan (RHS).
– cos q
2

170 / Optional Mathematics Class 10

b. LHS = cos4 pc + cos4 3pc + cos4 pc – 3pc + cos4 pc – pc
8 8 8 8

pc 3pc – cos 3pc 4 – cos pc 4
8 8 8 8
= cos4 + cos4 + +

= cos4 pc + cos4 3pc + cos4 3pc + cos4 pc
8 8 8 8

= 2 cos4 pc + 2 cos4 3pc
8 8

= 1 4 cos4 pc + 4 cos4 3pc
2 8 8

= 1 1 + cos 2 × pc 2 1 + cos 2 × 3pc 2
2 8 8
+

= 1 1 + cos pc 2 1 + cos 3pc 2
2 4 4
+

= 1 1+ 1 2 1– 1 2
2 2 2
+

= 1 2+1 2 2–1 2
2
+
22

= 1 ( 2)2 + 2. 2.1 + 12 + ( 2)2 – 2. 2.1 + 12
2 2

= 1 (2 + 1 + 2 + 1) = 1 × 6 = 3 (RHS) Trigonometry
4 2 2

Example 4: Find the value of sin 18º without using table or calculator.

Solution:

Let A = 18º, then 5A = 90º

or, 3A = 90º – 2A

or, cos 3A = cos (90º – 2A)

or, 4 cos3A – 3 cos A = sin 2A

or, cos A(4 cos2A – 3) = 2 sin A cos A

or, 4(1 – sin2A) – 3 = 2 sin A

or, 4 sin2A + 2 sin A – 1 = 0

∴ sin A = – 2 ± 4 – 4.4(– 1)
2×4

= – 2 ± 4 + 16 = –2± 20 = 2(– 1 ± 5)
8 8 8

i.e. sin 18º = – 1 ± 5
4

But the value of sin 18º is always positive. Hence, sin 18º = – 1+ 5.
4


Trigonometrical Ratios of Sub Multiple Angles / 171

Let us Memorize Sub-Multiple Angles
Multiple Angles

sin A = 2 sin A cos A
2 2
sin 2A = 2 sin A cos A
1. 2 tan A 2 cot A
= 2 tan A = 2 cot A 2 2
1 + tan2A 1 + cot2A = =
1 + cot2 A 1 + cot2 A
2 2

cos 2A = cos2A – sin2A cos A = cos2 A – sin2 A = 2 cos2 A – 1
2 2 2
A A
2. = 2 cos2A – 1 = 1 – 2 sin2A A 1 – tan2 2 cot2 2 – 1
= 1 – 2 sin2 2
= 1 – tan2A = cot2A – 1 = 1 + tan2 A = cot2 A + 1
1 + tan2A cot2A + 1 2 2

2 tan A 2 tan A
1 – tan2A 2
3. tan 2A = tan A =
A
1 – tan2 2

cot2A – 1 cot2 A –1
2 cot A 2
4. cot 2A = cot A = A
2
2 cot

5. sin 3A = 3 sin A – 4 sin3A sin A = 3 sin A – 4 sin3 A
6. cos 3A = 4 cos3A – 3 cos A 3 3

AA
cos A = 4 cos3 3 – 3 cos 3

3 tan A – tan3 A
1– 3 3
7. tan 3A = 3 tan A – tan3­A tan A =
1– 3 tan2A 3 A
tan2 3

cot3 A – 3 cot A
3 –1 3
8. cot 3A = cot3A – 3 cot A cot A =
3 cot2A –1 A
3 cot2 3

Exercise 5.2A

1. Find the value of sin θ, cos θ and tan θ when,

a. sin q = 35 b. cos q = 21 c. tan q = 5
2 2 2 12

2. Prove the following identities.

a. 1 +sincoqs q = cot q2 b. 1 tan q q = tan q
+ sec 2

172 / Optional Mathematics Class 10

c. 1 +sicnoAs2 A2+ sin A A2 d. 1 +sicnoAs A– – cos A A
+ cos 2 2
= tan A = cot
A sin 2

e. 11 ++ sin b – cos b = tan b2 f. 1 + cos a + sin a = cot a2
sin b+ cos b 1 – cos a + sin a

sin3 B – cos3 B 1 sin3 a + cos3 a 1
g. 2 – 2 2 sin 2 + 2 2
B B = 1 + sin B h. a a = 1 – sin a
sin 2 2 2 2
cos cos

i. cot A – tan A = 2 cot A j. cot b + tan b = 2 cosec b
2 2 2 2

3. Prove the following:

a. cos4 q – sin4 q = cos q b. cosec q– cot q = tan q
6 6 3 2
+
c. cosec A + cot A = cot A d. 2 sin a – sin 2a = cot2 a
2 2 sin a sin 2a 2

e. 22 ssiinn b – sin 2b = tan2 b2 f. 1 c+ossinq q = 1 + tan q
b+ sin 2b 1 – tan 2
q
2

g. 1 +sinco2sa2a · 1 cos a a = tan a2 h. 1 + sin q = cos q + sin q
+ cos 2 2

i. cos2 pc – q – sin2 pc – q = sin q2 j. 1 – 2 sin2 pc – A = sin A Trigonometry
4 4 4 4 4 2

k. 2sisnin2qq = cos2 q 1– tan2 q
2 2

4. a. If tan a = 21, find the value of tan α.
3

b. If cot b = 3, find the value of cot β.
2

Exercise 5.2B

1. a. If cos 30º = 23, find the values of sin 15º and cos 15º.

b. If cos 330º = 23, find the values of sin 165º and cos 165º.
2212°
c. If sin 45º = 1, prove that tan = 3–2 2.
2

2. Prove the following trigonometric identities:

a. tan 45° – A = 1 – sin A b. tan 45° + A = 1 + sin A
2 1 + sin A 2 1 – sin A

Trigonometrical Ratios of Sub Multiple Angles / 173

c. tan 45° + a + tan 45° – a = 2 sec a
2 2

d. cot 45° + q + tan 45° – q = 2 (sec q – tan q)
2 2
q
A + pc A – pc = 12+cosisnAA sin 2 – 1 + sin q q
2 4 2 4 1 + sin q 2
e. cot – tan f. cos q – = cot
2
pc A pc A = sin2A pc q pc q 1
g. sin2 8 + 2 – sin2 8 – 2 h. sin q sin 3 – 3 sin 3 + 3 = 4 sin q
3

i. sin q cos3 q + cos q sin3 q = 1 sin q j. cos6 a + sin6 a = 1 – 3 sin2a
2 2 2 2 2 2 2 4

k. cos q = 8 cos4 q – 8 cos2 q + 1
4 4

3. Show that:

a. sin4 pc + sin4 3pc + sin4 5pc + sin4 7pc = 3
8 8 8 8 2

b. cos2 pc + cos2 3pc + cos2 5pc + cos2 7pc = 2
8 8 8 8

c. 1 + cos pc 1 + cos 3pc 1 + cos 5pc 1 + cos 7pc = 1
8 8 8 8 8

d. (cos A – cos B)2 + (sin A+ sin B)2 = 4 sin2 A + B
2

e. (cos A + cos B)2 + (sin A – sin B)2 = 4 cos2 A+ B
2

4. a. If sin q = 35, prove that sin q = 111275.
3

b. If tan a = 51, prove that tan a = 5375.
3

c. If tan A = 43, prove that sin A = 2245.
2

d. If tan a = 1 – m tan b2, prove that cos β = 1c–osmac–osma.
2 1 + m

5. a. If cos q = 1 a + 1 , prove that cos q = 1 a2 + 1 .
2 2 2 2 a2

b. If cos a = 1 m + 1 , prove that cos a = 1 m3 + 1 .
3 2 m 2 m3

c. If sin b = 1 p + 1 , prove that sin b = – 1 p3 + 1 .
3 2 p 2 p3

6. a. If sin 18° = 5– 1 , then find the value of sin 36°.
4

b. If sin 15° = 3 –1 , then find the value of cos 30°.
3 2



174 / Optional Mathematics Class 10

5.3 Transformation of Product to Sum of Difference

Introduction

We have established the compound angle formulae of trigonometrical ratios as

sin A cos B + cos A sin B = sin (A + B) ... ... ... (1)

sin A cos B – cos A sin B = sin (A – B) ... ... ... (2)

Adding equations (1) and (2) we have,

2 sin A cos B = sin (A + B) + sin (A – B)

Again subtracting equation (2) from equation (1), we get

2 cos A sin B = sin (A + B) – sin (A – B)

Again, we have

cos A cos B + sin A sin B = cos (A – B) ... ... ... (3)

cos A cos B – sin A sin B = cos (A + B) ... ... ... (4)

Adding equations (3) and (4) we have,

2 cos A cos B = cos (A + B) + cos (A – B)

And subtracting equation (4) from equation (3), we have

2 sin A sin B = cos (A – B) – cos (A + B)

Example 1: Express the following products into sum or difference form. Trigonometry

a. sin 40° cos 28° b. cos 8θ cos 3θ

Solution:

a. We have, 2 sin A cos B = sin (A + B) + sin (A – B)

Hence, sin 40° cos 28°
= 21  [2 sin 40° cos 28°]
= 21 [sin (40° + 28°) + sin (40° – 28°)]
= 12  [  sin 68° + sin 12°]

b. We have, 2 cos A cos B = cos (A + B) + cos (A – B)
Hence, cos 8θ cos 3θ

= 12[2 cos 8θ cos 3q]
= 21[cos (8θ + 3θ) + cos (8θ – 3θ)]
= 12[ cos 11θ + cos 5q]

Transformation of Product to Sum of Difference / 175

Example 2: Prove that : 2 cos (45° + q) sin (45° – q) = 1 – sin 2q
Solution:

Here, LHS = 2 cos(45° + q) sin (45° – q)

= sin (45° + q + 45° – q) – sin {45° + q – (45° – q)}

= sin 90° – sin {45° + q – 45° + q}

= 1 – sin 2q (RHS)
Example 3: Prove that :

a. tan 50° – tan 40° = 2 tan 10°

b. cos 20° cos 40° cos 80° = 1
c. cos A cos (60° – A) cos 8
Solution: 1
(60° + A) = 4 cos 3A

b. LHS = tan 50° – tan 40°

= csoins 50° – sin 40° = 2[sin 50° cos 40° – cos 50° sin 40°]
50° cos 40° 2 cos 50° cos 40°

= cos 2 sin (50° – 40°) – 40°) = cos 2 sin 10° 10°
(50° + 40°) + cos (50° 90° + cos

= 2cosisn1100°° = 2 tan 10° (RHS)

b. LHS = cos 20° cos 40° cos 80°

= 21 (2 cos 20° cos 40°) cos 80°

= 21 [cos (20° + 40°) + cos (20° – 40°)] cos 80°

= 12 [cos 60° + cos 20°] cos 80°

= 21 × 1 cos 80° + 1 × 1 [2 cos 20° cos 80°]
2 2 2

= 41 cos 80° + 1 [cos (20° + 80°) + cos (20° – 80°)]
4

= 14 cos 80° + 1 [cos 100° + cos 60°]
4

= 14 cos 80° + 1 cos (180° – 80°) + 1 × 1
4 4 2

= 41 cos 80° – 1 cos 80° + 1 = 1 (RHS)
4 8 8

c. LHS = cos A cos (60° – A) cos (60° + A)

= 1 cos A [ 2 cos (60° – A) cos (60° + A)]
2

176 / Optional Mathematics Class 10

= 1 cos A [ cos (60° – A + 60° + A) + cos (60° – A – 60° – A)]
2

= 1 cos A [cos 120° + cos (– 2A)]
2

= 1 cos A – 1 + cos 2A
2 2

= 1 cos A – 1 + 2 cos2A – 1
2 2

= 1 cos A 2 cos2A – 3
2 2

= 1 [4 cos3A – 3 cos A]
4

= 1 cos 3A (RHS)
4

Example 7: Prove that : sin 36° sin 72° sin 108° sin 144° = 5
16

Solution:

LHS = sin 36° sin 72° sin 108° sin 144°

= 1 ( 2 sin 36° sin 144°) ( 2 sin 72° sin 108°)
4
1
= 4 [cos (36° – 144°) – cos (36° + 144°)] [cos (72° – 108°) – cos (72° + 108°)]

= 1 [cos 108° – cos 180°] [cos 36° – cos 180°] Trigonometry
4

= 1 [cos (90° + 18°) – (– 1)] [cos 36° – (– 1)]
4

= 1 [– sin 18° – (– 1)] [cos 36° – (– 1)]
4

= 1 – 5– 1 + 1 5+ 1 + 1
4 4 4

= 1 5– 5 5+ 5
4 4 4

= 1 × 25 – 5 = 5 (RHS).
4 4×4 16

Exercise 5.3A

1. Express the following products into sum or difference form:

a. 2 cos 48° sin 20° b. sin 40° sin 50° c. sin 105° cos 15°

d. 2 sin 7θ sin 3θ e. sin 4θ cos 2θ f. cos 75° cos 15°

2. Prove that:
a. 2 sin (45° + A) cos (45° – A) = 1 + sin 2A

b. 2 sin (45° + A) sin (45° – A) = cos 2A

c. 2 cos (45° + A) cos (45° – A) = cos 2A

Transformation of Product to Sum of Difference / 177

d. sin 75° cos 15° = 2 + 3
4

e. sin 15° sin 45° = 3–1
4

f. 2 cos 105° cos 15° + 1 = 0
2

g. sec 45° + A sec 45° – A = 2 sec A
2 2

h. cosec pc + q cosec pc – q = 2 sec 2q
4 4

Exercise 5.3B

1. Prove that:

a. tan 70° – 2 tan 50° = tan 20° b. tan 80° = 2 tan 70° + tan 10°

c. 2 tan 40° = tan 65° – tan 25° d. tan 235° – tan 215° = 2 tan 20°

2. Prove that:

a. 8 sin 10º sin 50º sin 70º = 1 b. 8 cos 10º cos 50º cos 70º = 3
c. cos 20º cos 40º cos 80º = 18
e. tan 20º tan 40º tan 80º = 3 d. sin 20º sin 40º sin 80º = 3
8

f. 8 cos 80° cos 140° cos 160° = 1

g. 4 cos q cos 2q cos 4q = cos 7q + cos 5q + cos 3q + cos q

h. 4 cos a cos 2a sin 4a = sin 7a + sin 5a + sin 3a + sin a

3. Prove the following identities:

a. sin α sin (60° – α) sin (60° + α) = 1 sin 3α
4

b. cos α cos (120° – α) cos (120° + α) = 1 cos 3α
4

c. sin q sin (120° + q) sin (120° – q) = 1 sin 3q
4

d. cos A cos 7pc + A cos 7pc – A = 1 cos 3A
3 3 4

4. Prove the following:

a. 16 cos 10º cos 30º cos 50º cos 70º = 3

b. 16 sin 10º sin 30º sin 50º sin 70º = 1

c. cos 2pc cos 4pc cos 8pc cos 14pc = 1
15 15 15 15 16

178 / Optional Mathematics Class 10

d. cos pc cos 3pc cos 5pc cos 7pc = 3
18 18 18 18 16

e. cos 18º cos 36º cos 72º cos 144º = – 1 cot 18º
16

f. sin 6º sin 42º sin 66º sin 78º = 1
16

g. tan 6º tan 42º tan 66º tan 78º = 1

h. cos 12º cos 24º cos 48º cos 84º = 1
16



5.4 Transformation of Sum of Difference to Product

Introduction

In previous section, we have used following formulae,

sin (A + B) + sin (A – B) = 2 sin A cos B ... ... ... (1)

sin (A + B) – sin (A – B) = 2 cos A sin B ... ... ... (2)

cos (A + B) + cos (A – B) = 2 cos A cos B ... ... ... (3)

cos (A – B) – cos (A + B) = 2 sin A sin B ... ... ... (4)

Let, A + B = C ... .... ... (5)

and A – B = D ... ... ... (6) Trigonometry

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

2A = C + D ⇒ A = C + D
2

Subtracting equation (6) from equation (5) we have,

2B = C – D ⇒ B = C – D
2

Now from (1), (2), (3) and (4), we have

sin C + sin D = 2 sin C + D cos C – D
2 2

sin C – sin D = 2 cos C + D sin C – D
2 2

cos C + cos D = 2 cos C + D cos C – D
2 2

cos D – cos C = 2 sin C + D sin C – D
2 2

Example 1: Express the following sum or difference into product form.

a. cos 20° – cos 70° b. sin 6θ – sin 2θ

Solution:

a. We have, cos D – cos C = 2 sin C + D sin C – D
2 2

Transformation of Sum of Difference to Product / 179

Hence, cos 20° – cos 70°

= 2 sin 20° + 70° sin 70° – 20°
2 2

= 2 sin 45° sin 25°

= 2 × 1 × sin 25°
2
1 × sin 25°
= 2× 2× 2

= 2 sin 25°

b. We have, sin C – sin D = 2 cos C + D sin C – D
2 2

Hence, sin 6θ – sin 2θ

= 2 cos 6q + 2q sin 6q – 2q
2 2

= 2 cos 4θ sin 2θ.

Example 2: Prove that:

a. cos 75° + cos 15° = 1 b. cos 10° + cos 110° + cos 130° = 0
Solution: sin 75° + sin 15°

a. LHS = cos 75° + cos 15°
sin 75° + sin 15°

2 cos 75° + 15° cos 75° – 15° 1
2 2
= = cos 45° cos 30° = 2 = 1 (RHS)
2 sin 75° + 15° cos 75° – 15° sin 45° cos 30° 1
2 2
2

b. LHS = cos 10° + cos 110° + cos 130°

= 2 cos 10° + 110° cos 10° – 110° + cos 130°
2 2

= 2 cos 60° cos (– 50°) + cos (180° – 50°)

= 2 × 1 cos 50° – cos 50°
2

= cos 50° – cos 50° = 0 (RHS)

Example 3: Prove that:

a. cos 7θ + cos 5θ + cos 3θ + cos θ = 4 cos θ cos 2θ cos 4θ
b.
cos a – cos 2a + cos 3a = cot 2α
c. sin a – sin 2a + sin 3a
d.
Solution: 2 cos pc cos 9pc + cos 3pc + cos 5pc = 0
13 13 13 13
3
cos2A + cos2(A – 120° ) + cos2(A + 120°) = 2

a. LHS = cos 7θ + cos 5θ + cos 3θ + cos θ

= 2 cos 7q + 5q cos 7q – 5q + 2 cos 3q + q cos 3q – q
2 2 2 2

180 / Optional Mathematics Class 10

= 2 cos 6θ cos θ + 2 cos 2θ cos θ

= 2 cos θ [cos 6θ + cos 2θ]

= 2 cos θ × 2 cos 6q + 2q cos 6q – 2q
2 2

= 4 cos θ cos 2θ cos 4θ (RHS)

b. LHS = cos a – cos 2a + cos 3a
sin a – sin 2a + sin 3a

= cos a + cos 3a – cos 2a
sin a + sin 3a – sin 2a

2 cos a + 3a cos 3a – a – cos 2a
2 2
=
a + 3a 3a – a
2 sin 2 cos 2 – sin 2a

= 2 cos 2a cos a – cos 2a
2 sin 2a cos a – sin 2a

= cos 2a (2 cos a – 1)
sin 2a (2 cos a – 1)

= cot 2α (RHS).

c. LHS = 2 cos pc cos 9pc + cos 3pc + cos 5pc
13 13 13 13

= 2 cos pc cos 9pc + 2 cos 3pc + 5pc cos 3pc – 5pc Trigonometry
13 13 13 × 2 13 × 2

= 2 cos pc cos 9pc + 2 cos 8pc cos – 2pc
13 13 26 26

= 2 cos pc cos 9pc + 2 cos 4pc cos pc
13 13 13 13

= 2 cos pc cos 9pc + cos 4pc
13 13 13

= 2 cos pc 2 cos 9pc + 4pc cos 9pc – 4pc
13 13 × 2 13 × 2

= 2 cos pc 2 cos pc cos 5pc
13 2 26

= 4 cos pc × 0 × cos 5pc
13 26

= 0 (RHS).

d. LHS = cos2A + cos2(A – 120°) + cos2(A + 120°)

= 12 [2 cos2A + 2 cos2(A – 120°) + 2 cos2(A + 120°)]
= 12 [1 + cos 2A + 1 + cos 2(A – 120°) + 1 + cos 2(A + 120°)]
= 12 [1 + cos 2A + 1 + cos (2A – 240°) + 1 + cos (2A + 240°)]

Transformation of Sum of Difference to Product / 181

= 21 3 + cos 2A + 2 cos 2A – 240° + 2A + 240° cos 2A – 240° – 2A – 240°
2 2

= 21 3 + cos 2A + 2 cos 2A cos (– 240°)

= 12 3 + cos 2A + 2 cos 2A cos  240°

= 12 3 + cos 2A + 2 cos 2A – 1
2

= 12 [3 + cos 2A – cos 2A] = 3 (RHS)
2

Example 4: Prove that:

a. sin 5A + sin 8A – sin 7A – sin 4A = cot 6A
b. cos 7A – cos 5A + cos 4A – cos 8A
Solution:
sin a sin2α – sin2b cos b = tan (a – b)
cos a – sin b

a. LHS = sin 5A + sin 8A – sin 7A – sin 4A
cos 7A – cos 5A + cos 4A – cos 8A

= sin 5A – sin 7A + sin 8A – sin 4A
cos 7A – cos 5A + cos 4A – cos 8A

2 cos 5A + 7A sin 5A – 7A + 2 cos 8A + 4A sin 8A – 4A
2 2 2 2
=
2 sin 7A + 5A sin 5A – 7A + 2 sin 4A + 8A sin 8A – 4A
2 2 2 2

= 2 cos 6A sin (– A) + 2 cos 6A sin 2A
2 sin 6A sin (– A) + 2 sin 6A sin 2A

= 2 cos 6A sin 2A – 2 cos 6A sin A
2 sin 6A sin 2A – 2 sin 6A sin A

= 2 cos 6A (sin 2A – sin A) = cos 6A = cot 6A (RHS)
2 sin 6A (sin 2A – sin A) sin 6A

b. LHS = sin a sin2α – sin2b cos b
cos a – sin b

= 2 sin 2 sin2α – 2 sin2b cos b
a cos a – 2 sin b

= 1 – cos 2α – (1 – cos 2b)
sin 2a – sin 2b

= cos 2b – cos 2a
sin 2a – sin 2b

2 sin 2b + 2a sin 2a – 2b
2 2
=
2 cos 2a + 2b sin 2a – 2b
2 2

= sin (b + α) sin (a – b) = tan (a + b) (RHS)
cos (a + b) sin (a – b)

182 / Optional Mathematics Class 10

Example 5: If sin x = k sin y, prove that tan x–y = k–1 tan x+y
2 k+1 2

Solution: sin x k
sin y 1
Here, sin x = k sin y i.e. =

By componendo and dividendo, we have,

ssiinn x – sin y = k – 1
x + sin y k + 1

2 cos x + y sin x – y
2 2
or, x + y x – y = k – 1
2 sin 2 cos 2 k + 1

or, cot x + y tan x – y = k – 1
2 2 k + 1

or, tan x – y = k – 1 1
2 k + 1
cot x + y
2

∴ tan x – y = k – 1 tan x + y
2 k + 1 2

Exercise 5.4A

1. Express the following sum or difference into product form: Trigonometry

a. cos 70° – cos 40° b. sin 70° – sin 20° c. sin 75° + sin 15°

d. cos 4θ – cos 2θ e. sin 7θ – sin θ f. sin 7b + cos 2b

2. Find the value of:

a. csoins 75° + csions1155°° b. csoins 8800°° +– scions2200°°
75° +

c. csoins 70° + sin 1700°° d. csoins 343400°° –– scions328900°°
10° – cos

3. Prove that:

a. csoins 80° – sin 20° = 13 b. csoins 7755°° +– scions1155°° = 1
80° + cos 20° 3

c. csoins 5q – cos 7q = tan 6θ d. csoins 79AA –– cos 9A = tan 8A
7q – sin 5q sin 7A

e. ssiinn 22aa + sin 2b = tan (a + bb)) f. ccooss qq + cos b = cot q + b cot b – q
– sin 2b tan (a – – cos b 2 2

g. csoins (p + 2)q – sin pq = cot (p + 1)θ h. cos 40° + sin 40° = cot 5°
pq – cos (p + 2)q cos 40° – sin 40°

Transformation of Sum of Difference to Product / 183

i. cos 38º cos 46º – sin 14º sin 22º = 1 cos 24º
2

j. 1 + cos 5pc cos 2pc = sin 5pc sin 2pc
7 7 7 7

k. cos248º – sin212º = 5+1
8

4. Prove the following identities:

a. sin 70° – sin 10° – sin 50° = 0 b. sin 35° + sin 25° – cos 5° = 0

c. cos 50° + sin 50° = 2 cos 5° d. sin 20° + cos 20° = 2 cos 25°

Exercise 5.4B

1. Prove the following trigonometric identities:

a. sin 4α cos 2α + cos 3α sin α = sin 5α cos α.

b. sin 11q sin q + sin 7q sin 3q = sin 2θ sin θ
4 4 4 4

c. sin 5θ – sin 3θ + 2 sin θ cos 2θ = 2 sin 2θ cos 3θ

h. 2 sin (45° + θ) cos (45° – θ) = 1 + sin 2θ

i. 2 sin 10° sin 40° = 3 – cos 50°
2

d. sin 7θ + 2 sin 2θ – sin 3θ = 2 sin 2θ (1 + cos 5θ)

e. cos 30° – cos 50° + cos 70° – cos 90° = 2 3 sin 10° cos 20°

f. sin 10° + sin 20° + sin 40° + sin 50° = 2 sin 75° cos 5°

g. sin A + sin 2A + sin 4A + sin 5A = 4 cos A cos 3A sin 3A
2 2

h. sin α + sin 3α + sin 5α + sin 7α = 4 cos α cos 2α sin 4α

i. sin2A + sin2(A – 120º) + sin2(A + 120º) = 3
2

2. Show that:

a. csoisn 2q + sin 5q – sin q = tan 2θ
2q + cos 5q + cos q

b. sin 5Aco+s 5A – cos A 3A = cot A
sin A–2 sin

c. csoins 3A – cos 2A + cos A = cot 2A
3A – sin 2A + sin A

d. csoins 7A – sin 6A + sin 5A = tan 6A
7A – cos 6A + cos 5A

184 / Optional Mathematics Class 10

e. ssiinn qq csoins 2q + sin 3q sin 6q = tan 5θ
2q + sin 3q cos 6q

f. ssiinn 6A cos 3A – sin 8A cos A = 1 2 tan A
4A sin 3A – cos 2A cos A – 2 tan2A

3. Prove the following identities:

a. sin A4+csoisnA3Aco+s sin 5A + sin 7A = tan 4A
2A cos 4A

b. csoisn 22AA++ sin 4A + sin 6A + sin 8A = tan 5A
cos 4A + cos 6A + cos 8A

c. ssiinnAAcsoins 2A + sin 3A sin 6A + sin 4A sin 13A = tan 9A
2A + sin 3A cos 6A + sin 4A cos 13A

d. csoisn 3A – cos 5A – cos A + cos 7A = – cot 2A
3A + sin 5A – sin A – sin 7A

e. 11 +– cos A + cos B – cos (A + B) = tan A cot B
cos A – cos B – cos (A + C) 2 2

f. ssiinn (A – C) + 2 sin A + sin (A + C) = sin A
(B – C) + 2 sin B + sin (B + C) sin B

g. ccooss3AA++22ccooss35AA++ccooss57AA = cos 2A – sin 2A tan 3A

h. csoins (a + b) – 2 sin a + sin (a – b) = tan α Trigonometry
(a + b) – 2 cos a + cos (a – b)

4. a. If cot b = k cot q, prove that cosec (q – b) = k+ 1 cosec (q + b)
k–1

b. If tan a = k tan b, prove that (k + 1) sin (a – b) = (k – 1) sin (a + b)

c. If cos x = k cos y, prove that cot x+y = 1+k tan x–y
2 1–k 2

d. If cos x + cos y = 1 and sin x + sin y = 14, prove that tan x + y = 43.
3 2

e. If cos x + cos y = 1 and sin x + sin y = 41, prove that tan x + y = 12.
2 2

f. If cos α = 3 and sin β = 1123, prove that: sin2  a – b = 1 and cos2  a + b = 1656.
5 2 65 2



Transformation of Sum of Difference to Product / 185

5.5 Conditional Trigonometric Identities

Introduction

Trigonometric identities such as sin2A+ cos2A = 1, tan2A = sec2A – 1, sin 2A = 2 sin A cos A etc. are
true for all the values of A. Some trigonometric identities, which are true for certain given condition are
known as conditional trigonometric identities. We will discuss the trigonometric identities for three given
angles A, B and C satisfying the condition A + B + C = 180°.
Here are some of the important results we often use to prove condtional identities.
If A + B + C = 180° (or pc), then A + B = 180° – C.
Now taking some trigonometric ratios on both sides, we have
a. sin (A + B) = sin (180° – C) = sin C ∴ sin (A + B ) = sin C
b. cos (A + B) = cos (180° – C ) = – cos C ∴ cos (A + B ) = – cos C
c. tan (A + B) = tan (180° – C ) = – tan C ∴ tan (A + B ) = – tan C

d. cot (A + B) = cot (180° – C ) = – cot C ∴ cot (A + B ) = – cot C
Example 1: If A + B + C = 180°, prove that tan A + tan B + tan C = tan A tan B tan C.
Solution:

Here, A + B + C = 180°

or, A + B = 180° – C

or, tan (A + B) = tan (180° – C)

or, tan A + tan B = – tan C
1 – tan A tan B

or, tan A + tan B = – tan C + tan A tan B tan C

or, tan A + tan B + tan C = tan A tan B tan C Proved.
Example 2: If A + B + C = pc, prove that cot A cot B + cot B cot C + cot C cot A = 1
Solution:

LHS = cot A cot B + cot B cot C + cot C cot A

= cot A cot B + cot C (cot B + cot A)

= cot A cot B – cot (A + B) (cot B + cot A) [ cot (A + B) = – cot C]

= cot A cot B – cot B cot A – 1 (cot B + cot A)
cot B + cot A

= cot A cot B – cot B cot A + 1

= 1 (RHS) Proved.

186 / Optional Mathematics Class 10

Alternatively,

As, A + B = pc – C

cot (A + B) = cot (pc – C)
or, ccoottBBc+otcAot–A1 = – cot C

or, cot A cot B – 1 = – cot B cot C – cot C cot A

∴ cot A cot B + cot B cot C + cot C cot A = 1 Proved.

Example 3: If A + B + C = 180º, prove that cos A – cos B + cos C = – 1 + 4 cos A sin B cos C2 .
2 2

Solution:

LHS = cos A – cos B + cos C

= 2 sin A + B sin B – A + cos C
2 2

Here, A + B = 90º – C
2 2 2

or, sin A + B = sin 90° – C = cos C
2 2 2

LHS = 2 cos C sin B – A + 2 cos2 C –1
2 2 2

= 2 cos C sin B – A + cos C –1
2 2 2
Trigonometry
= 2 cos C sin B – A + sin A + B –1
2 2 2 2 2

= – 1 + 2 cos C sin B cos A – cos B sin A + sin A cos B + cos A sin B
2 2 2 2 2 2 2 2 2

= – 1 + 2 cos C 2 sin B cos A
2 2 2

= – 1 + 4 cos A sin B cos C (RHS) Proved.
2 2 2
Example 4: If A + B + C = pc, prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

Solution:
LHS = sin 2A + sin 2B + sin 2C
= 2 sin (A + B) cos (A – B) + sin 2C
As, A + B = pc – C,
sin (A + B) = sin (pc – C) = sin C
LHS = 2 sin C cos (A – B) + 2 sin C cos C
Again, cos (A + B) = cos (pc – C) = – cos C
LHS = 2 sin C [cos (A – B) – cos (A + B)]
= 2 sin C [cos A cos B + sin A sin B – cos A cos B + sin A sin B]
= 2 sin C × 2 sin A sin B

= 4 sin A sin B sin C (RHS) Proved.

Conditional Trigonometric Identities / 187

Example 5: If A + B + C = 180º, prove that: cos2A + cos2B + cos2C = 1 – 2 cos A cos B cos C.

Solution:

LHS = cos2A + cos2B + cos2C

= 1 + cos 2A + 1 + cos 2B + cos2C
2 2

= 1 + 1 + 1 (cos 2A + cos 2B) + cos2C
2 2 2

= 1 + 1 [2 cos (A + B) cos (A – B)] + cos2C
2

As, A + B = 180º – C,

or, cos (A + B) = cos (180º – C) = – cos C
LHS = 1 – cos C cos (A – B) + cos2C
= 1 – cos C [cos (A – B) – cos C]
= 1 – cos C [cos (A – B) + cos (A + B)]
= 1 – cos C [cos A cos B + sin A sin B + cos A cos B – sin A sin B]
= 1 – cos C × 2 cos A cos B

= 1 – 2 cos A cos B cos C [RHS] Proved.

Example 6: If A + B + C = pc, show that: cos A – cos B + cos C = 4 cos pc + A cos pc – B cos pc + C
2 2 2 4 4 4
Solution:

LHS = cos A – cos B + cos C
222
A + B B – A pc C
= 2 sin 4 sin 4 + sin 2 – 2

= 2 sin A + B sin B – A + sin 2 pc – C
4 4 4
A + B B – A pc – C pc – C
= 2 sin 4 sin 4 + 2 sin 4 cos 4

= 2 sin A + B sin B – A + 2 sin A + B cos pc – C
4 4 4 4
A + B B – A pc – C
= 2 sin 4 sin 4 + cos 4

= 2 sin A + B sin B – A + sin pc – pc – C
4 4 2 4
A + B B – A 2pc – pc + C
= 2 sin 2 sin 4 + sin 4

= 2 sin A + B sin B – A + sin pc + C
4 4 4
A + B B – A + pc + C B – A – pc – C
= 2 sin 4 2 sin 8 cos 8

= 2 sin A + B 2 sin 2B + 2C cos – 2A – 2C
4 8 8
A + B B + C A + C
= 2 sin 4 2 sin 4 cos 4

= 4 sin A + B sin B + C cos C + A
4 4 4
pc A + B pc B + C pc – B
= 4 cos 2 – 4 cos 2 – 4 cos 4

188 / Optional Mathematics Class 10

= 4 cos 2pc – A – B cos 2pc – B – C cos pc – B
4 4 4
pc + C pc + A pc – B
= 4 cos 4 cos 4 cos 4

= 4 cos pc + A cos pc – B cos pc + C [RHS] Proved.
4 4 4

Example 7: If A + B + C = pc, show that: cos 3A + cos 3B – cos 3C = – 1 – 4 cos 3A cos 3B sin 3C
2 22

Solution:

LHS = cos 3A + cos 3B – cos 3C

= 2 cos 3A + 3B cos 3A – 3B – cos 3C
2 2
3A + 3B 3pc – 3C
As, A + B = pc – C, we have 2 = 2

So, cos 3A + 3B = cos 3pc – 3C = – sin 3C
2 2 2 2

LHS = – 2 sin 3C cos 3A – 3B – 1 – 2 sin232C
2 2

= – 2 sin 3C cos 3A – 3B – sin 3C –1
2 2 2

= – 2 sin 3C cos 3A – 3B + cos 3A + 3B –1
2 2 2

= – 2 sin 3C cos 3A – 3B + cos 3A + 3B –1
2 2 2 2 2

= – 2 sin 3C cos 3A cos 3B + sin 3A sin 3B + cos 3A cos 3B – sin 3A sin 3B – 1 Trigonometry
2 2 2 2 2 2 2 2 2

= – 1 – 2 sin 3C × 2 cos 3A cos 3B
2 2 2

= – 1 – 4 cos 3A cos 3B sin 3C (RHS) Proved.
2 2 2

Example 8: If α + β + γ = p2c, prove that: sin2α + sin2β + sin2γ = 1 – 2 sin α sin β sin γ.
Solution:

LHS = sin2α + sin2β + sin2γ

= 1 [2 sin2α + 2 sin2b] + sin2γ
2
1
= 2 [1 – cos 2α + 1 – cos 2b] + sin2γ

= 1 [2 – (cos 2α + cos 2β)] + sin2γ
2
1
= 1 – 2 × 2 cos (α + β) cos (α – β) + sin2γ

Here, α + β + γ = pc so, cos (α + β) = cos pc – g = sin γ
2 2
= 1 – sin γ cos (α – β) + sin2γ

= 1 – sin g {cos (α – β) – sin g}

= 1 – sin g {cos (α – β) – cos (α + β)}

= 1 – sin g {cos α cos β + sin α sin β – cos α cos β + sin α sin b}

= 1 – 2 sin α sin β sin γ (RHS) Proved.

Conditional Trigonometric Identities / 189

Exercise 5.5

1. If A + B + C = 180°, prove that:

a. tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

b. tan B tan C + tan C tan A + tan A tan B = 1
2 2 2 2 2 2

c. cot A + cot B + cot C = cot A cot B cot C
2 2 2 2 2 2

d. tan (A + B – C) + tan (B + C – A) + tan (C + A – B) = tan 2A tan 2B tan 2C

2. If A + B + C = pc, prove that:

a. sin A + sin B + sin C = 4 cos A cos B cos C
2 2 2

b. sin A – sin B + sin C = 4 sin A cos B sin C
2 2 2

c. cos A + cos B + cos C = 1 + 4 sin A sin B sin C
2 2 2

d. cos A + cos B – cos C = – 1 + 4 cos A cos B sin C
2 2 2

3. If A + B + C = pc, prove that:

a. sin 2A – sin 2B + sin 2C = 4 cos A sin B cos C

b. sin 2A + sin 2B – sin 2C = 4 cos A cos B sin C

c. cos 2A + cos 2B + cos 2C = – 1 – 4 cos A cos B cos C

d. cos 2A – cos 2B + cos 2C = 1 – 4 sin A cos B sin C
4. If A + B + C = pc, prove that:
a. sin2A + sin2B + sin2C = 2 + 2 cos A cos B cos C
b. sin2A + sin2B – sin2C = 2 sin A sin B cos C
c. cos2A + cos2B – cos2C = 1 – 2 sin A sin B cos C
d. sin22A + sin22B + sin22C = 2 – 2 cos 2A cos 2B cos 2C
e. cos22A + cos22B + cos22C = 1 + 2 cos 2A cos 2B cos 2C
5. If A + B + C = 180°, prove that:

a. sin2 A + sin2 B + sin2 C = 1 – 2 sin A sin B sin C
2 2 2 2 2 2

b. sin2 A – sin2 B + sin2 C = 1 – 2 cos A sin B cos C
2 2 2 2 2 2

c. cos2 A + cos2 B + cos2 C = 2 + 2 sin A sin B sin C
2 2 2 2 2 2

d. cos2 A + cos2 B – cos2 C = 2 cos A cos B sin C
2 2 2 2 2 2

190 / Optional Mathematics Class 10

6. If A, B and C are the angles of a triangle, prove that:

a. cos A + cos B + cos C = 4 cos B + C cos C + A cos A + B
2 2 2 4 4 4
pc – A pc – B pc – C
= 4 cos 4 cos 4 cos 4

b. sin A + sin B + sin C = 1 + 4 sin pc – A sin pc – B sin pc – C
2 2 2 4 4 4
A B C pc + A pc + B pc – C
c. cos 2 + cos 2 – cos 2 = 4 cos 4 cos 4 cos 4

d. tan A + tan B + C = sec A sec B + C
2 2 2 2

7. If A, B and C are the angles of a triangle, prove that:

a. sin 3A + sin 3B + sin 3C = – 4 cos 3 A cos 3 B cos 3 C
2 2 2

b. cos 4A + cos 4B + cos 4C = – 1 + 4 cos 2A cos 2B cos 2C

c. sin 5A + sin 5B + sin 5C = 4 cos 5A cos 5B cos 5C
2 2 2

d. sin 6A + sin 6B + sin 6C = 4 sin 3A sin 3B sin 3C

8. If A + B + C = p2c, prove that:
a. tan A tan B + tan B tan C + tan C tan A = 1

b. cot A + cot B + cot C = cot A cot B cot C

c. sin2A + sin2B + sin2C = 1 – 2 sin A sin B sin C

d. sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C Trigonometry

e. ssiinn 22AA + sin 2B + sin 2C = cot A cot B
+ sin 2B – sin 2C

9. If A, B and C are angles of a triangle, prove that:

a. sincBossAin C + cos B A + cos C B = 2
sin C sin sin A sin

b. cossBincAos C + sin B A + sin C B = 2 tan A tan B tan C
cos C cos cos A cos

10. a. If α + β = θ, then show that: cos2α + cos2β + cos2θ = 1 + 2 cos α cos β cos θ

b. If α + β + g = 2k, prove that sin k sin (k – a) + sin (k – b) sin (k – g) = sin b sin g
c. If cos A + cos B + cos C = 0, prove that:
cos 3A + cos 3B + cos 3C = 12 cos A cos B cos C.
11. If A + B + C = pc, prove that:

a. cos (A + B – C) + cos (B + C – A) + cos (C + A – B) = 1 + 4 cos A cos B cos C

b. sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C

c. cos A sin B sin C + sin A cos B sin C + sin A sin B cos C – cos A cos B cos C = 1

d. sin (B + 2C) + sin (C + 2A) + sin (A + 2B) = 4 sin B–C sin C–A sin A–B
2 2 2


Conditional Trigonometric Identities / 191

5.6 Trigonometric Equations

Introduction

There are two types of equality of expressions. They are
a. Indentity b. Equation
An identity is true for every value of the variable where as the equation is true for only certain values
of the variable.
Consider the following equalities involving trigonometrical ratios,
sin x cosec x = 1 is an identity where as cos2x = 1 is an equation.
Let us take cos2x + sin2x = 1.
When x = 0°, then cos20° + sin20° = 1.
When x = 30°, then cos230° + sin230° = 1.
When x = 45°, then cos245° + sin245° = 1.
Hence, cos2x + sin2x = 1 is true for x = 0°, 30°, 45°, ... ... and every value of x.
Therefore, cos2x + sin2x = 1 is an identity.
Now, take sin x + cos x = 1.
For x = 0°, then sin 0° + cos 0° = 1
For x = 30°, then sin 30° + cos 30° ≠ 1
For x = 45°, then sin 45° + cos 45° ≠ 1
For x = 90°, then sin 90° + cos 90° = 1

Hence, sin x + cos x = 1 is true for x = 0°, 90°, ... ... but not for x = 30°, 45°, .... ... .

Therefore, sin x + cos x = 1 is an equation.
Now, consider an equation cos x = 21.
The value of the unknown angle x in above equation which satisfy the given equation is called
solution of the equation.
We know that the trigonometrical ratio of a certain angle has unique value. But the value of angle
when the value of its trigonometrical ratio is given is not unique. Study the graph of y = cos x.

Y

360° 720°

y = cos x

ValuesX' O 60° 90° 270° 300° 420° 450° 630° 660° X

180° 540°

Y' Angles

192 / Optional Mathematics Class 10

From the graph, when y = cos x = 12c,otshineeviaslu12e.sTohfexnaurme b6e0r°,o3f0ro0o°,ts42o0r °s,o6lu6Yt0io°,n90.s.°.o.f..a.nHyetnricgeo,nthoemreetarrice
infinite number of values of angle whose
equation depends on the given range.

Methods of solving trigonometric equations 2nd Quadrant 1st Quadrant

90° + θ 360° + θ

Use the base equations. 180° – θ 90° – θ

S A

a. If sin A = sin a then A = a, Sin/cosec postive All postive 360°
b. If cos A = cos b, then A = b, X
c. If tan A = tan q, then A = q etc. 180° Cos/sec postive
Some special angles and their values in range 0° – 360°. X'
If sin q = 0, then q = 0°, 180°, 360° C
Tan/cot postive
4th Quadrant
T
270° + θ
3rd Quadrant 360° – θ

180° + θ
270° – θ

If cos q = 0, then q = 90°, 270° Y' 270°
If tan q = 0, then q = 0°, 180°, 360°
If sin q = 21, then q = 30°, 150°
If tan q = 1, then q = 45°, 225°
If cos q = 21, then q = 60°, 300°
If sin q = – 1, than q = 270° If tan q = 13, then q = 30°, 210°
If sin q = 12, then q = 45°, 135°
If cos q = – 1, then q = 180°

Example 1: Solve : sin 3A = cos 2A (0° ≤ x ≤ 90°) Trigonometry

Solution:

Here, sin 3A = cos 2A
or, sin 3A = sin (90° – 2A)
or, 3A = 90° – 2A
or, 3A + 2A = 90°
or, 5A = 90°
\ A = 18°
Hence, A = 18°

Example 2: Solve : sin2x = sin x (0° ≤ x ≤ 180°)

Solution:

sin2x = sin x

or, sin2x – sin x = 0

or, sin x (sin x – 1) = 0

either sin x = 0 ... ... ... (1)

or, sin x – 1 = 0 ... ... ... (2)

From (1), sin x = 0

or, sin x = sin 0° or sin 180°

∴ x = 0°, 180°

Trigonometric Equations / 193

From (2), sin x – 1 = 0

or, sin x = sin 90°

∴ x = 90°

Hence, x = 0°, 90°, 180°.
Example 3: Solve : 2 sin2x = 3 cos x (0° ≤ x ≤ 360°)

Solution:

2 sin2x = 3 cos x

or, 2 (1 – cos2x) – 3 cos x = 0

or, 2 – 2 cos2x – 3 cos x = 0

or, 2 cos2x + 3 cos x – 2 = 0

or, 2 cos2x + 4 cos x – cos x – 2 = 0

or, 2 cos x (cos x + 2) – 1 (cos x + 2) = 0

or, (cos x + 2) (2 cos x – 1) = 0

Either, cos x + 2 = 0 ... ... ... (1)

or, 2 cos x – 1 = 0 ... ... ... (2)

From (1), cos x = – 2 which is not possible as cos x can not be less than – 1.

This equation has no solution.

From (2), 2 cos x – 1 = 0

or, cos x = 1
2
or, cos x = cos 60°

∴ x = 60°

Again, cos x = cos (360° – 60°)

or, cos x = cos 300°

∴ x = 60°, 300°.
Example 4: Solve : 7 sin2x + 3 cos2x = 4 (0° ≤ x ≤ 360°)

Solution:

We have, 7 sin2x + 3 cos2x = 4 Taking negative sign,

or, 7 sin2x + 3 (1 – sin2x) = 4 sin x = – 1 = – sin 30°
2
or, 7 sin2x + 3 – 3 sin2x = 4
or, sin x = sin (180° + 30°)
or, 4 sin2x = 1
or, sin x = sin 210°

or, sin2x = 1 ∴ x = 210°
4
1 and, sin x = – sin 30°
or, sin x = ± 2
or, sin x = sin (360° – 30°)

Taking positive sign, or, sin x = sin 330°

sin x = 1 ∴ x = 330°
2

194 / Optional Mathematics Class 10

or, sin x = sin 30°

∴ x = 30°

sin x = sin (180° – 30°)

or, sin x = sin 150°

∴ x = 150°

Hence, x = 30°, 150°, 210°, 330°
Example 5: Solve : 2 cos θ = cot θ (0° ≤ θ ≤ 360°)

Solution:

2 cos θ = cot θ

or, 2 cos θ = cos θ
sin θ

or, 2 sin θ cos θ = cos θ

[ Be careful ! do not cancel cos θ]

or, 2 sin θ cos θ – cos θ = 0

or, cos θ (2 sin θ – 1) = 0

Either, cos θ = 0 ... ... ... (1)
... ... ... (2)
or, 2 sin θ – 1 = 0

From (1), cos θ = 0 = cos 90° or cos 270°

∴ θ = 90°, 270° Trigonometry

From (2), 2 sin θ – 1 = 0

or, sin θ = 1
2

or, sin θ = sin 30°

∴ θ = 30°

And, sin θ = sin (180° – 30°)

or, sin θ = sin 150°

∴ θ = 150°

Hence, θ = 30°, 90°, 150°, 270°

Example 6: Solve : 3 cos θ + sin θ = 1 (0° ≤ θ ≤ 360°)

Solution:

3 cos θ + sin θ = 1
or, 3 cos θ = 1 – sin θ
Squaring both sides,
or, 3 cos2θ = 1 – 2 sin θ + sin2θ
or, 3 – 3 sin2θ = 1 – 2 sin θ + sin2θ
or, 4 sin2θ – 2 sin θ – 2 = 0
or, 2 sin2θ – sin θ – 1 = 0

Trigonometric Equations / 195

or, 2 sin2θ – 2 sin θ + sin θ – 1 = 0
or, 2 sin θ (sin θ – 1) + 1 (sin θ – 1) = 0

or, (sin θ – 1) (2 sin θ + 1) = 0

Either, sin θ – 1 = 0 ... ... ... (1)

or, 2 sin θ + 1 = 0 ... ... ... (2)
From (1), sin θ = 1
or, sin θ = sin 90°
∴ θ = 90°.

From (2), 2 sin θ = – 1

or, sin θ = – 1
2

or, sin θ = – sin 30°

or, sin θ = sin (180° + 30°)

∴ θ = 210°

Again, sin θ = sin (360° – 30°)

or, sin θ = sin 330°

∴ θ = 330°

Hence, θ = 90°, 210°, 330°

Checking, when θ = 90°

LHS = 3 × cos 90° + sin 90° = 0 + 1 = 1 = RHS

When θ = 210°,

LHS = 3 × cos 210° + sin 210° = 3×– 3 – 1 = – 2 ≠ RHS
2 2
When θ = 330°,

LHS = 3 × cos 330° + sin 330° = 3× 3 – 1 = 1 = RHS.
2 2
As, θ = 210° does not satisfy given equation, the required solution is θ = 90°, 330°.

Note: Generally when we solve a trigonometrical equations by squaring both sides or by another means,
some solutions obtained may not satisfy the given equation. So we must check each solution by
substituting the solution obtained in the equation. Reject those solutions which do not satisfy the
given equation.

Alternatively,

3 cos θ + sin θ = 1 (0° ≤ θ ≤ 360°)

Dividing both sides by (coeff. of x)2 + (coeff. of y)2

i.e. ( 3)2 + 12 = 2

i.e. 3 cos θ + sin θ = 1
2 2

196 / Optional Mathematics Class 10