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Prove the following: 2(sin6θ + cos6θ) – 3(sin4θ + cos4θ) + 1 = 0

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Question

Prove the following:

2(sin6θ + cos6θ) – 3(sin4θ + cos4θ) + 1 = 0

Sum
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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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Chapter 2: Trigonometry - 1 - MISCELLANEOUS EXERCISE - 2 [Page 33]

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Balbharati Mathematics and Statistics 1 (Arts and Science) [English] Standard 11 Maharashtra State Board
Chapter 2 Trigonometry - 1
MISCELLANEOUS EXERCISE - 2 | Q 10) vi) | Page 33

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