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Question
Prove the following:
3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6x + cos6x) = 13
Solution
(sin x – cos x)4
= [(sin x – cos x)2]2
= (sin2x + cos2x – 2 sin x cos x)2
= (1 – 2 sin x cos x)2
= 1 – 4 sin x cos x + 4 sin2x cos2x
(sin x + cos x)2
= sin2x + cos2x + 2 sin x cos x
= 1 + 2 sin x cos x
sin6x + cos6x
= (sin2x)3 + (cos2x)3
= (sin2x + cos2x)3 – 3 sin2x cos2x (sin2x + cos2x) ...[∵ a3 + b3 = (a + b)3 – 3ab (a + b)]
= 13 – 3 sin2x cos2x (1)
= 1 – 3 sin2x cos2x
L.H.S. = 3 (sin x – cos x)4 + 6 (sin x + cosx)2 + 4 (sin6x + cos6x)
= 3 (1 – 4 sin x cos x + 4 sin2x cos2x) + 6 (1 + 2 sin x cos x) + 4 (1 – 3 sin2x cos2x)
= 3 – 12 sin x cos x + 12 sin2x cos2x + 6 + 12 sin x cos x + 4 – 12 sin2x cos2x
= 13
= R.H.S.
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