Problem 1 :
prove that
sin θ/2 sin 7θ/2 + sin 3θ/2 sin 11θ/2 = sin2θ sin 5θ.
Solution :
L.H.S :
= sin θ/2 sin 7θ/2 + sin 3θ/2 sin 11θ/2
= (1/2) [2 sin θ/2 sin 7θ/2] + (1/2) [2 sin3θ/2 sin 11θ/2]
2 sin θ/2 sin 7θ/2 = cos 3θ - cos 4θ
2 sin3θ/2 sin 11θ/2 = cos 4θ - cos 7θ
Applying the those values, we get
= (1/2) [cos 3θ - cos 4θ] + (1/2) [cos 4θ - cos 7θ]
= (1/2) [cos 3θ - cos 4θ + cos 4θ - cos 7θ]
= (1/2) [cos 3θ - cos 7θ]
= (1/2) [2 sin 5θ sin 2θ]
= sin 5θ sin 2θ
Hence proved.
Problem 2 :
Prove that
cos (30°−A) cos (30°+A) + cos (45°−A) cos (45°+A) = cos2A + (1/4)
Solution :
L.H.S
= cos (30°−A) cos (30°+A) + cos (45°−A) cos (45°+A)
= (1/2) (2cos (30°−A) cos (30°+A)) + (1/2) (2cos (45°−A) cos (45°+A))
= (1/2) [cos 60 + cos 2A + cos 90 + cos 2A]
= (1/2) [(1/2) + cos 2A + 0 + cos 2A]
= (1/2) [(1/2) + 2cos 2A]
= 1/4 + cos 2 A
Hence proved.
Problem 3 :
Prove that
(sin x + sin3x + sin5x + sin7x) / (cos x + cos3x + cos5x + cos7x) = tan4x.
Solution :
= (sin x + sin3x + sin5x + sin7x) / (cos x + cos3x + cos5x + cos7x)
sin x + sin3x = 2 sin 2x cos x -------(1)
sin5x + sin7x = 2 sin 6x cos x-------(2)
cos x + cos3x = 2 cos 2x cos x -------(3)
cos5x + cos7x = 2 cos 6x cos x -------(4)
(1) + (2)
= 2 sin 2x cos x + 2 sin 6x cos x
= 2 cos x (sin 2x + sin 6x) -----(A)
(3) + (4)
= 2 cos 2x cos x + 2 cos 6x cos x
= 2 cos x (cos 2x + cos 6x) -----(B)
(A)/(B)
= 2 cos x (sin 2x + sin 6x) / 2 cos x (cos 2x + cos 6x)
= (sin 2x + sin 6x) / (cos 2x + cos 6x)
Again using the formula for sin C + sin D and cos C + cos D, we get
= 2 sin 4x cos 2x / 2 cos 4x cos 2x
= tan 4x
Hence proved.
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