हिंदी

Prove that 2(sin^6A + cos^6A) – 3(sin^4A + cos^4A) + 1 = 0.

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प्रश्न

Prove that 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0.

प्रमेय
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उत्तर

sin6A + cos6A = (sin2A)3 + (cos2A)3

 = (1 – cos2A)3 + (cos2A)3    ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`

= 1 – 3 cos2A + 3(cos2A)2 – (cos2A)3 + cos6A   ...[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]

= 1 – 3 cos2A(1 – cos2A) – cos6A + cos6A

= 1 – 3 cos2A sin2A

sin4A + cos4A = (sin2A)2 + (cos2A)2

= (1 – cos2A)2 + (cos2A)2

= 1 – 2 cos2A + (cos2A)2 + (cos2A)2    ...[∵ (a – b)2 = a2 – 2ab + b2]

= 1 – 2 cos2A + 2 cos4A

= 1 – 2 cos2A(1 – cos2A)

= 1 – 2 cos2A sin2A

L.H.S. = 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1

= 2(1 – 3 cos2A sin2A) – 3(1 – 2 cos2A sin2A) + 1

= 2 – 6 cos2A sin2A – 3 + 6 cos2A sin2A + 1

= 0

= R.H.S.

∴ 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0

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अध्याय 6: Trigonometry - Q.4
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