Question

Prove that :
$(si{n}^8\theta -co{s}^8\theta )=(si{n}^2\theta -co{s}^2\theta )(1-2si{n}^2\theta co{s}^2\theta )$

Answer 1

We have ,

$(si{n}^8\theta -co{s}^8\theta )=(si{n}^4\theta {)}^2-(co{s}^4\theta {)}^2=(si{n}^4\theta -co{s}^4\theta )(si{n}^4\theta +co{s}^4\theta )$

$\Rightarrow LHS=(si{n}^2\theta -co{s}^2\theta )(si{n}^2\theta +co{s}^2\theta )(si{n}^4\theta +co{s}^4\theta )$

$\Rightarrow LHS=(si{n}^2\theta -co{s}^2\theta )\left[\right(si{n}^2\theta {)}^2+(co{s}^2\theta {)}^2+2si{n}^2\theta co{s}^2\theta -2si{n}^2\theta co{s}^2\theta ]$

$\Rightarrow LHS=(si{n}^2\theta -co{s}^2\theta )\left[\right(si{n}^2\theta +co{s}^2\theta {)}^2-2si{n}^2\theta co{s}^2\theta ]$

$\Rightarrow LHS=(si{n}^2\theta -co{s}^2\theta )(1-2si{n}^2\theta co{s}^2\theta )=RHS$