Question

evaluate sin^6+cos^6

Answer 1

er, sin6ө + cos6ө =1-3sin2өcos2ө
(sin2ө)3 + (cos2ө)3 = (sin2ө + cos2ө)3 − 3 (sin2ө cos2ө)(sin2ө + cos2ө)   [since a + b = (a+b)3 − 3ab(a+b)]
                           = 1 − 3sin2өcos2ө    [Since sin2ө + cos2ө = 1]

Answer 2

cos^6 x+sin^6 x=(cos^2x)^3+(sin^2x)^3
                            =(cos^2x+sin^2x)(cos^4x+sin^4x-cos2x.sin2x)
                            =1*((cos^2x+sin^2x)^2-2*cos^2x.sin^2x-sin^2x.cos^2x
                            =1-3sin^2x cos^2x.