Question

ii.
Prove that Sin^6theta + Cos^6theta = 1-3Sin^2theta .Cos^2theta​

Answer 1

Answer:

Given: sin6ө + cos6ө =1-3sin2өcos2ө

Consider LHS

=sin6ө + cos6ө

=(sin2ө)3 + (cos2ө)3  

= (sin2ө + cos2ө)3 − 3 (sin2ө cos2ө)(sin2ө + cos2ө)        [a3 +b3+ b = (a+b)3 − 3ab(a+b)]

= 1 − 3sin2ө.cos2ө             [sin2ө + cos2ө = 1]

=RHS

Hence Proved.