Question

(sin theta )^8 - (cos theta )^8=(sin^2 thetan- cos ^2theta) (1-2sin^2theta)

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).

hope this will help you.

thank you......