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# RD Chapter 8- Transformation Formulae Ex-8.1 Interview Questions Answers

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Question 1 :
Express each of the following as the sum or difference of sines and cosines:
(i) 2 sin 3x cos x
(ii) 2 cos 3x sin 2x
(iii) 2 sin 4x sin 3x
(iv) 2 cos 7x cos 3x

(i) 2 sin 3x cos x
By using the formula,
2 sin A cos B = sin (A + B) + sin (A – B)
2 sin 3x cos x = sin (3x + x) + sin (3x – x)
= sin (4x) + sin (2x)
= sin 4x + sin 2x

(ii) 2 cos 3x sin 2x
By using the formula,
2 cos A sin B = sin (A + B) – sin (A – B)
2 cos 3x sin 2x = sin (3x + 2x) – sin (3x – 2x)
= sin (5x) – sin (x)
= sin 5x – sin x

(iii) 2 sin 4x sin 3x
By using the formula,
2 sin A sin B = cos (A – B) – cos (A + B)
2 sin 4x sin 3x = cos (4x – 3x) – cos (4x + 3x)
= cos (x) – cos (7x)
= cos x – cos 7x

(iv) 2 cos 7x cos 3x
By using the formula,
2 cos A cos B = cos (A + B) + cos (A – B)
2 sin 3x cos x = cos (7x + 3x) + cos (7x – 3x)
= cos (10x) + cos (4x)
= cos 10x + cos 4x

Question 2 :
Prove that:
(i) 2 sin 5π/12 sin π/12 = 1/2
(ii) 2 cos 5π/12 cos π/12 = 1/2
(iii) 2 sin 5π/12 cos π/12 = (√3 + 2)/2

(i) 2 sin 5π/12 sin π/12 = 1/2
By using the formula,
2 sin A sin B = cos (A – B) – cos (A + B)
2 sin 5π/12 sin π/12 = cos (5π/12 – π/12) – cos (5π/12 + π/12)
= cos (4π/12) – cos (6π/12)
= cos (π/3) – cos (π/2)
= cos (180o/3) – cos (180o/2)
= cos 60° – cos 90°
= 1/2 – 0
= 1/2
Hence Proved.
(ii) 2 cos 5π/12 cos π/12 = 1/2
By using the formula,
2 cos A cos B = cos (A + B) + cos (A – B)
2 cos 5π/12 cos π/12 = cos (5π/12 + π/12) + cos (5π/12 – π/12)
= cos (6π/12) + cos (4π/12)
= cos (π/2) + cos (π/3)
= cos (180o/2) + cos (180o/3)
= cos 90° + cos 60°
= 0 + 1/2
= 1/2
Hence Proved.
(iii) 2 sin 5π/12 cos π/12 = (√3 + 2)/2
By using the formula,
2 sin A cos B = sin (A + B) + sin (A – B)
2 sin 5π/12 cos π/12 = sin (5π/12 + π/12) + sin (5π/12 – π/12)
= sin (6π/12) + sin (4π/12)
= sin (π/2) + sin (π/3)
= sin (180o/2) + sin (180o/3)
= sin 90° + sin 60°
= 1 + √3
= (2 + √3)/2
= (√3 + 2)/2
Hence Proved.

Question 3 :

showthat:

(i) sin 50o cos 85o =(1 – √2sin 35o)/2√2

(ii) sin25o cos 115o = 1/2 {sin 140o – 1}

(i) sin 50o cos85o = (1 – √2sin 35o)/2√2

By using the formula,

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

sin A cos B = [sin (A +B) + sin (A – B)] / 2

sin 50o cos 85o =[sin(50o + 85o) + sin (50o –85o)] / 2

= [sin (135o) + sin (-35o)]/ 2

= [sin (135o) – sin (35o)]/ 2 (since, sin (-x) = -sin x)

= [sin (180o – 45o)– sin 35o] / 2

= [sin 45o – sin 35o]/ 2

= [(1/√2) – sin 35o] / 2

= [(1 – sin 35o)/2] / 2

= (1 – sin 35o) / 22

Hence proved.

(ii) sin 25o cos115o = 1/2 {sin 140o – 1}

By using the formula,

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

sin A cos B = [sin (A +B) + sin (A – B)] / 2

sin 20o cos 115o =[sin(25o + 115o) + sin (25o –115o)] / 2

= [sin (140o) + sin (-90o)]/ 2

= [sin (140o) – sin (90o)]/ 2 (since, sin (-x) = -sin x)

= 1/2 {sin 140o – 1}

Hence proved.

Question 4 :
Prove that:
4 cos x cos (π/3 + x) cos (π/3 – x) = cos 3x

Let us consider LHS:

4 cos x cos (π/3 + x) cos (π/3 – x) = 2 cos x (2 cos(π/3 + x) cos (π/3 – x))

By using the formula,

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

2 cos x (2 cos (π/3+x) cos (π/3 – x)) = 2 cos x (cos(π/3+x + π/3-x) + cos (π/3+x – π/3+ x))

= 2 cos x (cos (2π/3) + cos (2x))

= 2 cos x {cos 120° + cos 2x}

= 2 cos x {cos (180° – 60°) + cos 2x}

= 2 cos x (cos 2x – cos 60°) (since, {cos (180° – A) =– cos A})

= 2 cos 2x cos x – 2 cos x cos 60°

= (cos (x + 2x) + cos (2x – x)) – (2cos x)/2

= cos 3x + cos x – cos x

= cos 3x

= RHS

Hence Proved.

krishan