Question

Prove that sin6 θ + cos6 θ + 3sin2 θ cos2 θ = 1

Answer 1

Answer:

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]