Problem 1 :
Show that
(cosθ-cos3θ)(sin8θ+sin2θ) / (sin5θ-sinθ)(cos4θ-cos6θ)=1
Solution :
cosθ - cos3θ = 2 sin 2θ sin θ -------(1)
sin8θ + sin2θ = 2 sin 5θ cos 3θ -------(2)
sin5θ - sinθ = 2 cos 3θ sin 2θ -------(3)
cos4θ - cos6θ = 2 sin 5θ sin θ -------(4)
(1) ⋅ (2) :
= 4 sin 2θ sin θ sin 5θ cos 3θ ------(A)
(3) ⋅ (4) :
= 4 cos 3θ sin 2θ sin 5θ sin θ ------(B)
(A)/(B) :
= 4 sin2θsinθsin 5θ cos 3θ / 4 cos 3θ sin 2θ sin 5θ sin θ
= 1
Hence proved.
Problem 2 :
Prove that :
sin x + sin2x + sin3x = sin2x (1 + 2cos x)
Solution :
sin x + sin2x + sin3x :
= sin 2x + 2 sin 2x cos x
= sin 2x (1 + 2 cos x)
Problem 3 :
Prove that
(sin 4x + sin2x) /(cos 4x + cos2x) = tan3x.
Solution :
sin 4x + sin2x = 2 sin 3x cos x --------(1)
cos 4x + cos2x = 2 cos 3x cos x --------(2)
(1) / (2)
= 2 sin 3x cos x / 2 cos 3x cos x
= tan 3x
Problem 4 :
Prove that
1 + cos2x + cos4x + cos6x = 4cos x cos 2x cos 3x.
Solution :
1 + cos2x + cos4x + cos6x :
= 1 + cos2x + 2 cos 5x cos x
= 2 cos2 x + 2 cos 5x cos x
= 2 cos x (cos x + cos 5x)
= 2 cos x (2 cos 3x cos 2x)
= 4 cos x cos 2x cos 3x
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