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cos^(6)A+sin^(6)A=1-k sin^(2)(2A)rArr k=...

cos^(6)A+sin^(6)A=1-k sin^(2)(2A)rArr k=

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cos^(6)A+sin^(6)A=1-3/4sin^(2)(2A)

cos^(6)A+sin^(6)A=1-3/4sin^(2)(2A)

cos^(6) A + sin^(6) A = 1 - k sin^2 (2A) implies k =

cos^6A-sin^6A=cos2A(1-1/4sin^2 2A)

Prove that: cos^(6)A-sin^(6)A=cos2A(1-(1)/(4)sin^(2)2A)

sin^(6)A + cos^(6)A + 3sin^(2)A.cos^(2)A=

sin ^ (6) A + cos ^ (6) A = 1-3sin ^ (2) A cos ^ (2) A

cos ^ (6) A + sin ^ (6) A = (1) / (4) (1 + 3cos ^ (2) 2A) cos6A + sin A = (1 + 3cos22A)

Prove the following : cos^6A-sin^6A = cos2A(1-1/4 sin^2 2A)

sin^(6)A+cos^(6)A+3sin^(2)A cos^(2)A=0 b.1c2d.3