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Prove that sin^6A + cos^6A = 1 – 3sin^2A . cos^2A.

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प्रश्न

Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.

प्रमेय
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उत्तर

L.H.S. = sin6A + cos6A

= (sin2A)3 + (cos2A)3   

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

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

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

= 1 – 3 cos2A sin2A

= R.H.S.

∴ sin6A + cos6A = 1 – 3sin2A . cos2A

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अध्याय 6: Trigonometry - Exercise

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