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• RD Sharma Class 11 Solutions for Maths

# Class 11 RD Sharma Solutions – Chapter 8 Transformation Formulae – Exercise 8.1

### (i) 2 sin 3θ cos θ

Solution:

By using the trigonometric identity,

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

Taking A = 3θ and B = θ

2 sin 3θ cos θ = sin (3θ+θ) + sin (3θ-θ)

= sin 4θ + sin 2θ

### (ii) 2 cos 3θ sin 2θ

Solution:

By using the trigonometric identity,

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

Taking A = 2θ and B = 3θ

2 cos 3θ sin 2θ = sin (3θ+2θ) + sin (2θ-3θ)

= sin 5θ + sin (-θ)

= sin 5θ – sin θ

### (iii) 2 sin 4θ sin 3θ

Solution:

By using the trigonometric identity,

2 sin A sin B = cos (A-B) – cos (A+B)

Taking A = 4θ and B = 3θ

2 sin 4θ sin 3θ = cos (4θ-3θ) – cos (4θ+3θ)

= cos θ – cos 7θ

### (iv) 2 cos 7θ cos 3θ

Solution:

By using the trigonometric identity,

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

Taking A = 7θ and B = 3θ

2 cos 7θ cos 3θ = cos (7θ+3θ) + cos (7θ-3θ)

= cos 10θ – cos 4θ

### (i)

Solution:

By using the trigonometric identity,

2 sin A sin B = cos (A-B) – cos (A+B)

Taking A =  and B =

Hence, LHS = RHS

### (ii)

Solution:

By using the trigonometric identity,

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

Taking A =  and B =

Hence, LHS = RHS

### (iii)

Solution:

By using the trigonometric identity,

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

Taking A =  and B =

= sin  + sin

= 1 +

Hence, LHS = RHS

### (i) sin 50° cos 85° =

Solution:

By using the trigonometric identity,

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

sin A cos B = (sin (A+B) + sin (A-B))

Taking A = 50° and B = 85°

sin 50° cos 85° = (sin (50°+85°) + sin (50°-85°))

(sin (135°) + sin (-35°))

(sin (180°-45°) – sin (35°))     (sin(-θ)=-sin θ)

(sin (45°) – sin (35°))     (sin(π-θ)=sin θ)

Hence, LHS = RHS

### (ii) sin 25° cos 115° =

Solution:

By using the trigonometric identity,

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

sin A cos B = (sin (A+B) + sin (A-B))

Taking A = 25° and B = 115°

sin 25° cos 115° = (sin (25°+85°) + sin (25°-115°))

(sin (140°) + sin (-90°))

(sin (180-40°) – sin (90°))     (sin(-θ)=-sin θ)

(sin (40°) – sin (90°))     (sin(π-θ)=sin θ)

(sin (40°) – 1)

Hence, LHS = RHS

### Question 4. Prove that :

Solution:

By using the trigonometric identity,

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

Taking A = +θ and B =

Using the identity again, we have

Taking A = 2θ and B = θ

Hence, LHS = RHS

### (i) cos 10° cos 30° cos 50° cos 70° =

Solution:

cos 10° cos 30° cos 50° cos 70° = cos 30° cos 10° cos 50° cos 70°

(cos 10° cos 50°) cos 70°

By using the trigonometric identity,

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

cos A cos B = [cos (A+B) + cos (A-B)]

Taking A = 10° and B = 50°

([cos (10°+50°) + cos (10°-50°)]) cos 70°

(cos (60°) + cos (-40°)) cos 70°

( + cos (40°)) cos 70°

cos 70° +  (cos 70° cos (40°))

Again using the identity, we get

cos 70° +  ([cos (70°+40°) + cos (70°-40°)])

cos 70° +  [cos (110°) + cos (30°)]

cos 70° +  [cos (110°) + ]

cos 70° +  cos (110°) +

(cos 70° + cos (110°)) +

(cos 70° + cos (180°-70°)) +

(cos 70° – cos (70°)) +

Hence, LHS = RHS

### (ii) cos 40° cos 80° cos 160° =

Solution:

cos 40° cos 80° cos 160° = cos 80° (cos 40° cos 160°)

By using the trigonometric identity,

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

cos A cos B = [cos (A+B) + cos (A-B)]

Taking A = 160° and B = 40°

= cos 80° ([cos (160°+40°) + cos (160°-40°)])

= cos 80° ([cos (200°) + cos (120°)])

= cos 80° ([cos (180°+20°) + cos (180°-60°)])

= cos 80° ([- cos (20°) + (-cos (60°))])

= cos 80° ([- cos (20°) – cos (60°)])

= cos 80° ([- cos (20°) – ])

(cos 80° cos (20°) +  cos 80°])

Again using the identity, we get

Hence, LHS = RHS

### (iii) sin 20° sin 40° sin 80° =

Solution:

sin 20° sin 40° sin 80° = (sin 20° sin 40°) sin 80°

By using the trigonometric identity,

2 sin A sin B = cos (A-B) – cos (A+B)

sin A sin B = [cos (A-B) – cos (A+B)]

Taking A = 40° and B = 20°

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

sin 80° [cos (20°) – cos (60°)]

sin 80° [cos (20°) – ]

[sin 80° cos (20°) –  sin 80°]

By using the trigonometric identity,

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

sin A cos B = [sin (A+B) + sin (A-B)]

Taking A = 80° and B = 20°

Hence, LHS = RHS

### (iv) cos 20° cos 40° cos 80° =

Solution:

cos 20° cos 40° cos 80° = cos 40° (cos 20° cos 80°)

By using the trigonometric identity,

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

cos A cos B = [cos (A+B) + cos (A-B)]

Taking A = 80° and B = 20°

Again using the identity, we get

Hence, LHS = RHS

### (v) tan 20° tan 40° tan 60° tan 80° = 3

Solution:

tan 20° tan 40° tan 60° tan 80° = tan 60°

By using the trigonometric identity,

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

cos A cos B = [cos (A+B) + cos (A-B)]

and, 2 sin A sin B = cos (A-B) – cos (A+B)

sin A sin B = [cos (A-B) – cos (A+B)]

Taking A = 40° and B = 20°

Again using the identity, we get

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

sin A cos B = [sin (A+B) + sin (A-B)]

Taking A = 80° and B = 20°

Hence, LHS = RHS

### (vi) tan 20° tan 30° tan 40° tan 80° = 1

Solution:

tan 20° tan 30° tan 40° tan 80° = tan 30°

By using the trigonometric identity,

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

cos A cos B = [cos (A+B) + cos (A-B)]

and, 2 sin A sin B = cos (A-B) – cos (A+B)

sin A sin B = [cos (A-B) – cos (A+B)]

Taking A = 40° and B = 20°

Again using the identity, we get

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

sin A cos B = \frac{1}{2}[sin (A+B) + sin (A-B)]

Taking A = 80° and B = 20°

Hence, LHS = RHS

### (vii) sin 10° sin 50° sin 60° sin 70° =

Solution:

sin 10° sin 50° sin 60° sin 70° = sin 60° (sin 10° sin 50° sin 70°)

= \frac{\sqrt{3}}{2} (sin (90-80°) sin (90-40°) sin (90-20°))

(cos (80°) cos (40°) cos (20°))

cos 40° (cos 80° cos 20°)

By using the trigonometric identity,

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

cos A cos B = [cos (A+B) + cos (A-B)]

Taking A = 80° and B = 20°

Again using the identity, we get

Hence, LHS = RHS

### (viii) sin 20° sin 40° sin 60° sin 80° =

Solution:

sin 20° sin 40° sin 60° sin 80° = sin 60° (sin 20° sin 40° sin 80°)

(sin 20° sin 40°) sin 80°

By using the trigonometric identity,

2 sin A sin B = cos (A-B) – cos (A+B)

sin A sin B = [cos (A-B) – cos (A+B)]

Taking A = 40° and B = 20°

By using the trigonometric identity,

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

sin A cos B = [sin (A+B) + sin (A-B)]

Taking A = 80° and B = 20°

Hence, LHS = RHS

### (i) sin A sin (B-C) + sin B sin (C-A) + sin C sin (A-B) = 0

Solution:

By using the trigonometric identity,

2 sin θ sin Φ = cos (θ-Φ) – cos (θ+Φ)

sin θ sin Φ = [cos (θ-Φ) – cos (θ+Φ)]

sin A sin (B-C) + sin B sin (C-A) + sin C sin (A-B) = ([cos (A-(B-C)) – cos (A+(B-C))]) + ([cos (B-(C-A)) – cos (B+(C-A))]) + ([cos (C-(A-B)) – cos (C+(A-B))])

(cos (A-B+C)) – cos (A+B-C) + cos (B-C+A) – cos (B+C-A) + cos (C-A+B) – cos (C+A-B))

(cos (A-B+C)) – cos (C+A-B) – cos (A+B-C) + cos (B-C+A) – cos (B+C-A) + cos (C-A+B))

= 0

Hence, LHS = RHS

### (ii) sin (B-C) cos (A-D) + sin (C-A) cos (B-D) + sin (A-B) cos (C-D) = 0

Solution:

By using the trigonometric identity,

2 sin θ cos Φ = sin (θ+Φ) + sin (θ-Φ)

sin θ cos Φ[sin (θ+Φ) + sin (θ-Φ)]

sin (B-C) cos (A-D) + sin (C-A) cos (B-D) + sin (A-B) cos (C-D) = ([sin (B-C+(A-D)) + sin (B-C-(A-D))]) + ([sin (C-A+(B-D)) + sin (C-A-(B-D))]) +([sin (A-B+(C-D)) + sin (A-B-(C-D))])

= ([sin (A+B-C-D) + sin (-A+B-C+D)]) + ([sin (-A+B+C-D) + sin (-A-B+C+D)]) +([sin (A-B+C-D) + sin (A-B-C+-D)])

(sin (A+B-C-D) + sin (-(A-B+C-D)) + sin (-(A-B-C+D)) + sin (-(A+B-C-D)) +sin (A-B+C-D) + sin (A-B-C+-D))

=(sin (A+B-C-D) – sin(A-B+C-D) – sin (A-B-C+D) – sin (A+B-C-D) +sin (A-B+C-D) + sin (A-B-C+-D))

= 0

Hence, LHS = RHS

### Question 7. Prove that : tan θ tan (60°-θ) tan (60°+θ) = tan 3θ

Solution:

tan θ tan (60°-θ) tan (60°+θ) = tan θ (tan (60°-θ)) (tan (60°+θ))

By using the trigonometric identity,

tan (a+b) =

tan (a+b) =

= tan 3θ

Hence, LHS = RHS

### Question 8. If α + β = 90°, show that the maximum value of cos(α) cos(β) is

Solution:

cos(α) cos(β) = y

By using the trigonometric identity,

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

cos A cos B = [cos (A+B) + cos (A-B)]

Taking A = α and B = β

cos(α) cos(β) = [cos (α+β) + cos (α-β)]

As, α + β = 90°

y = [cos (90°) + cos (α-β)]

y = [0 + cos (α-β)]

y = (cos (α-β))

AS, we know that range of cos function is [-1,1]

Hence, the maximum value of cos(α) cos(β) is

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