Question

prove that, sin^6θ+cos^6θ= 1-3sin^2θ cos^2θ.

Answer 1

Answer:

L.H.S. =sin6θ+cos6θ

=(sin2θ)3+(cos2θ)3

Put sin2θ=a and cos2θ=b

∴ L.H.S. =a3+b3

=(a+b)3−2ab(a+b)

=(sin2θ−cos2θ)3−3sin2θcos2θ(sin2θ+cos2θ)

=(1)3−3sin2θcos2θ[∵sin2θ+cos2θ=1]

=1−3sin2θcos2θ

= R.H.S.

Answer 2

Answer:

Step-by-step explanation: