Question

If sin$\theta$ + cos$\theta$ = x, prove that $si{n}^6\theta +co{s}^6\theta =\frac{4-3(x^2-1{)}^2}{4}$

Answer 1

$\sin \theta +\cos \theta =x$

$\Rightarrow {\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =x$

$\Rightarrow 1+2\sin \theta \cos \theta =x$

$\Rightarrow \sin \theta \cos \theta =(x-\frac{1}{2})$

${\sin }^6\theta +{\cos }^6\theta =({\sin }^2\theta {)}^3+({\cos }^2\theta {)}^3=({\sin }^2\theta +{\cos }^2\theta )({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta )$

$=\left(\right({\sin }^2\theta +{\cos }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta )=1-3{\sin }^2\theta {\cos }^2\theta$

$=1-\frac{3(x-1{)}^2}{4}$

$=\frac{4-3(x-1{)}^2}{4}$.