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