Question

I want to prove that

sin6ө + cos6ө=1-3sin2өcos2ө

prove it

Answer 1

Consider, 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]