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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 3Î¸ cos Î¸

Solution:

By using the trigonometric identity,

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

Taking A = 3Î¸ and B = Î¸

2 sin 3Î¸ cos Î¸ = sin (3Î¸+Î¸) + sin (3Î¸-Î¸)

= sin 4Î¸ + sin 2Î¸

### (ii) 2 cos 3Î¸ sin 2Î¸

Solution:

By using the trigonometric identity,

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

Taking A = 2Î¸ and B = 3Î¸

2 cos 3Î¸ sin 2Î¸ = sin (3Î¸+2Î¸) + sin (2Î¸-3Î¸)

= sin 5Î¸ + sin (-Î¸)

= sin 5Î¸ – sin Î¸

### (iii) 2 sin 4Î¸ sin 3Î¸

Solution:

By using the trigonometric identity,

2 sin A sin B = cos (A-B) – cos (A+B)

Taking A = 4Î¸ and B = 3Î¸

2 sin 4Î¸ sin 3Î¸ = cos (4Î¸-3Î¸) – cos (4Î¸+3Î¸)

= cos Î¸ – cos 7Î¸

### (iv) 2 cos 7Î¸ cos 3Î¸

Solution:

By using the trigonometric identity,

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

Taking A = 7Î¸ and B = 3Î¸

2 cos 7Î¸ cos 3Î¸ = cos (7Î¸+3Î¸) + cos (7Î¸-3Î¸)

= cos 10Î¸ – cos 4Î¸

### (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 50Â° cos 85Â° =

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 = 50Â° and B = 85Â°

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

(sin (135Â°) + sin (-35Â°))

(sin (180Â°-45Â°) – sin (35Â°))     (sin(-Î¸)=-sin Î¸)

(sin (45Â°) – sin (35Â°))     (sin(Ď€-Î¸)=sin Î¸)

Hence, LHS = RHS

### (ii) sin 25Â° cos 115Â° =

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 = 25Â° and B = 115Â°

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

(sin (140Â°) + sin (-90Â°))

(sin (180-40Â°) – sin (90Â°))     (sin(-Î¸)=-sin Î¸)

(sin (40Â°) – sin (90Â°))     (sin(Ď€-Î¸)=sin Î¸)

(sin (40Â°) – 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 = +Î¸ and B = -Î¸

Using the identity again, we have

Taking A = 2Î¸ and B = Î¸

Hence, LHS = RHS

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

Solution:

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

(cos 10Â° cos 50Â°) cos 70Â°

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 = 10Â° and B = 50Â°

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

(cos (60Â°) + cos (-40Â°)) cos 70Â°

( + cos (40Â°)) cos 70Â°

cos 70Â° +  (cos 70Â° cos (40Â°))

Again using the identity, we get

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

cos 70Â° +  [cos (110Â°) + cos (30Â°)]

cos 70Â° +  [cos (110Â°) + ]

cos 70Â° +  cos (110Â°) +

(cos 70Â° + cos (110Â°)) +

(cos 70Â° + cos (180Â°-70Â°)) +

(cos 70Â° – cos (70Â°)) +

Hence, LHS = RHS

### (ii) cos 40Â° cos 80Â° cos 160Â° =

Solution:

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

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 = 160Â° and B = 40Â°

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

= cos 80Â° ([cos (200Â°) + cos (120Â°)])

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

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

= cos 80Â° ([- cos (20Â°) – cos (60Â°)])

= cos 80Â° ([- cos (20Â°) – ])

(cos 80Â° cos (20Â°) +  cos 80Â°])

Again using the identity, we get

Hence, LHS = RHS

### (iii) sin 20Â° sin 40Â° sin 80Â° =

Solution:

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

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 = 40Â° and B = 20Â°

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

sin 80Â° [cos (20Â°) – cos (60Â°)]

sin 80Â° [cos (20Â°) – ]

[sin 80Â° cos (20Â°) –  sin 80Â°]

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 = 80Â° and B = 20Â°

Hence, LHS = RHS

### (iv) cos 20Â° cos 40Â° cos 80Â° =

Solution:

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

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 = 80Â° and B = 20Â°

Again using the identity, we get

Hence, LHS = RHS

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

Solution:

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

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 = 40Â° and B = 20Â°

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 = 80Â° and B = 20Â°

Hence, LHS = RHS

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

Solution:

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

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 = 40Â° and B = 20Â°

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 = 80Â° and B = 20Â°

Hence, LHS = RHS

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

Solution:

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

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

(cos (80Â°) cos (40Â°) cos (20Â°))

cos 40Â° (cos 80Â° cos 20Â°)

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 = 80Â° and B = 20Â°

Again using the identity, we get

Hence, LHS = RHS

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

Solution:

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

(sin 20Â° sin 40Â°) sin 80Â°

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 = 40Â° and B = 20Â°

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 = 80Â° and B = 20Â°

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 Î¸ sin Î¦ = cos (Î¸-Î¦) – cos (Î¸+Î¦)

sin Î¸ sin Î¦ = [cos (Î¸-Î¦) – cos (Î¸+Î¦)]

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 Î¸ cos Î¦ = sin (Î¸+Î¦) + sin (Î¸-Î¦)

sin Î¸ cos Î¦[sin (Î¸+Î¦) + sin (Î¸-Î¦)]

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 Î¸ tan (60Â°-Î¸) tan (60Â°+Î¸) = tan 3Î¸

Solution:

tan Î¸ tan (60Â°-Î¸) tan (60Â°+Î¸) = tan Î¸ (tan (60Â°-Î¸)) (tan (60Â°+Î¸))

By using the trigonometric identity,

tan (a+b) =

tan (a+b) =

= tan 3Î¸

Hence, LHS = RHS

### Question 8. If Î± + Î˛ = 90Â°, show that the maximum value of cos(Î±) cos(Î˛) is

Solution:

cos(Î±) cos(Î˛) = 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 = Î± and B = Î˛

cos(Î±) cos(Î˛) = [cos (Î±+Î˛) + cos (Î±-Î˛)]

As, Î± + Î˛ = 90Â°

y = [cos (90Â°) + cos (Î±-Î˛)]

y = [0 + cos (Î±-Î˛)]

y = (cos (Î±-Î˛))

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

Hence, the maximum value of cos(Î±) cos(Î˛) is

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