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Question
Prove the following:
2(sin6θ + cos6θ) – 3(sin4θ + cos4θ) + 1 = 0
Solution
sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ)3 – 3sin2θ cos2θ (sin2θ + cos2θ) ...[∵ a3 + b3 = (a + b)3 – 3ab(a + b)]
= (1)3 – 3 sin2θ cos2θ(1)
= 1 – 3 sin2θ cos2θ
sin4θ + cos4θ
= (sin2θ)2 + (cos2θ)2
= (sin2θ + cos2θ)2 – 2sin2θ cos2θ ...[∵ a2 + b2 = (a + b)2 – 2ab]
= 1 – 2sin2θ cos2θ
L.H.S. = 2(sin6θ + cos6θ) – 3(sin4θ + cos4θ) + 1
= 2(1 – 3 sin2θ cos2θ) – 3(1 – 2 sin2θ cos2θ) + 1
= 2 – 6 sin2θ cos2θ – 3 + 6 sin2θ cos2θ + 1
= 0
= R.H.S.
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