SUM TO PRODUCT TRIGONOMETRIC IDENTITES

Example 1 :

Express

sin 4A + sin 2A

in the form of product.

Solution :

Given expression sin 4A + sin 2A exactly matches with

sin C+ sin D = 2 sin (C+D)/2 cos (C-D)/2

Here C  =  4A and D  =  2A

sin 4A + sin 2A  =  2 sin (4A+2A)/2 cos (4A-2A)/2

=  2 sin (6A/2) cos (2A/2)

=  2 sin 3A cos A

Example 2 :

Express

sin 5A - sin 3A

in the form of product.

Solution :

Given expression sin 5A - sin 3A exactly matches with

sin C - sin D   2 cos (C+D)/2 sin (C-D)/2

Here C  =  5A and D  =  3A

sin 5A - sin 3A  =  2 cos (5A+3A)/2 sin (5A-3A)/2

=  2 sin (8A/2) cos (2A/2)

=  2 sin 4A cos A

Example 3 :

Express

cos 3A + cos 7A

in the form of product.

Solution :

Given expression cos 3A + cos 7A exactly matches with

cos C + cos D  =  2 cos (C+D)/2 cos (C-D)/2

Here C  =  3A and D  =  7A

cos 3A + cos 7A  =  2 cos (3A+7A)/2 cos (3A-7A)/2

=  2 sin (10A/2) cos (-4A/2)

=  2 sin 5A cos(-2A)

=  2 sin 5A cos 2A

Example 4 :

Evaluate

cos 15° - cos 75°

Solution :

cos C - cos D  =  -2 sin (C+D)/2 sin (C- D)/2

cos 15° - cos 75°  =  -2 sin (15°+75°)/2 sin (15°-75°)/2

=  -2 sin (90°)/2 sin (-60°)/2

=  -2 sin 45° sin (-30°)

=  2 (1/√2)(-1/2)

=  -1/√2

Example 5 :

Evaluate

sin 75° + sin 15°

Solution :

sin C + sin D  =  2 sin (C+D)/2 cos (C- D)/2

sin 75° + sin 15°  =  2 sin (75°+15°)/2 cos (75°-15°)/2

=  2 sin (90°)/2 cos (60°)/2

=  2 sin 45° cos (30°)

=  2 (1/√2)(√3/2)

=  √3/√2

Example 6 :

Prove that 

(cos 4t - cos 2t)/(sin 4t + sin 2t)  =  - tant

Solution :

(cos 4t - cos 2t)/(sin 4t + sin 2t)

cos 4t - cos 2t  =  -2 sin (4t+2t)/2 sin (4t-2t)/2

cos 4t - cos 2t  =  -2 sin 3t sin t  ----(1)

sin 4t + sin 2t  =  2 sin (4t+2t)/2 cos (4t-2t)/2

sin 4t + sin 2t  =  2 sin3t cost ---(2)

(1)/(2)

=  -2 sin 3t sin t / 2 sin3t cost

=  -sint/cost

=  - tant

Hence proved

Example 7 :

Simplify the given expression.

(cos 3A - cos 11A) / (sin 9A + sin 5A)

Solution :

= (cos 3A cos 11A) / (sin 9A + sin 5A)

cos C - cos D = -2 sin (C + D)/2 sin (C - D)/2

C = 3A and D = 11A

= -2 sin (3A + 11A)/2 sin (3A - 11A)/2

= -2 sin (14A/2) sin (- 8A/2)

= 2 sin 7A sin 4A -------(1)

sin C + sin D = 2 sin (C + D)/2 cos (C - D)/2

Here C = 9A and D = 5A

sin 9A + sin 5A = 2 sin (9A + 5A)/2 cos (9A - 5A)/2

= 2 sin (14A/2) cos (4A/2)

= 2 sin 7A cos 2A -------(2)

(1)/(2)

= (2 sin 7A sin 4A) / (2 sin 7A cos 2A)

= sin 4A/cos 2A

= sin 2(2A)/cos 2A

= 2sin 2A cos 2A / cos 2A

= 2 sin 2A

So, the answer is 2 sin 2A.

Example 8 :

 Simplify the given expression.

(cos4x - cos6x) / (sin 6x - sin4x)

Solution :

= (cos 4x - cos 6x) / (sin 6x - sin 4x)

cos C - cos D = -2 sin (C + D)/2 sin (C - D)/2

C = 4x and D = 6x

= -2 sin (4x + 6x)/2 sin (4x - 6x)/2

= -2 sin (10x/2) sin (- 2x/2)

= 2 sin 5x sin x -------(1)

sin C - sin D = 2 cos (C + D)/2 sin (C - D)/2

Here C = 6x and D = 4x

sin 6x - sin4x = 2 sin (6x + 4x)/2 cos (6x - 4x)/2

= 2 sin (10x/2) cos (2x/2)

= 2 sin 5x cos x -------(2)

(1)/(2)

= (2 sin 5x sin x) / (2 sin 5x cos x)

= sin x/cos x

= tan x

Example 9 :

Complete the identity

sin θ [sin(4θ) + sin 6θ)] =

Solution :

= sin θ [sin 4θ + sin 6θ] ----(1)

sin C + sin D = 2 sin (C + D)/2 cos (C - D)/2

Here C = 4θ and D = 6θ

sin 4θ + sin 6θ = 2 sin (4θ + 6θ)/2 cos (4θ - 6θ)/2

= 2 sin (10θ/2) cos (- 2θ/2)

= 2 sin (5θ) cos (- θ)

= 2 sin 5θ cos θ

By applying the above, we get

= sin θ [2 sin 5θ cos θ]

= sin 5θ (2 sin θ cos θ)

= sin 5θ sin θ

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