Question

Sin8 theta - cos8 theta = (sin2 theta - cos2 theta)(1-2sin2 theta + cos2 theta)Prove the above identity

Answer 1

sin⁸θ-cos⁸θ
=(sin⁴θ)²-(cos⁴θ)²
=(sin⁴θ+cos⁴θ)(sin⁴θ-cos⁴θ)
={(sin²θ)²+(cos²θ)²}{(sin²θ)²-(cos²θ)²}
={(sin²θ+cos²θ)²-2sin²θcos²θ}{(sin²θ+cos²θ)(sin²θ-cos²θ)}
={(1)²-2sin²θcos²θ}{(1)(sin²θ-cos²θ)} [∵, sin²θ+cos²θ=1]
=(sin²θ-cos²θ)(1-2sin²θcos²θ) (Proved)

Answer 2

now u can change 1=sin^2theta +cos^2theta.