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Sin6A+cos6A=1-3sin2Acos2A

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Posted by Palak Gupta 4 years, 10 months ago

Gaurav Seth 4 years, 10 months ago

a n s w e r
sin^6A + cos^6A

= (sin^2A)3 + (cos^2A)3

= (sin^2 A + cos^2 A) (sin^4 A – sin^2A cos^2A + cos^4A)

= 1 (sin^4A – sin^2Acos^2A + cos^4A + 2sin^2Acos^2A – 2sin^2Acos^2A)

= (sin2A + cos2A) – 3sin2A cos2A

= 1 – 3sin^2Acos^2A

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