Question

If $\sin x+{\sin }^{2}x=1$, then write the value of ${\cos }^{8}x+2{\cos }^{6}x+{\cos }^{4}x$.

Answer 1

${\cos }^{8}x+2{\cos }^{6}x+{\cos }^{4}x$

$\sin x=1-{\sin }^{2}x$

$\sin x={\cos }^{2}x$

${{(\cos }^{2}x)}^4+2{\cos }^{2}x({{\cos }^{2}x)}^2+{{(\cos }^{2}x)}^2$

$={{(\sin }^{2}x)}^2+2{\sin }^{2}x\times {\cos }^{2}x+{{(\cos }^{2}x)}^2$ $[{(a+b)}^2=a^2+b^2+2ab]$

$={({\sin }^{2}x+{\cos }^{2}x)}^2$

$=1^2$

$=1$