Question

If $\sin \theta +{\sin }^2\theta +{\sin }^3\theta =1$, prove that ${\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta =4$

Answer 1

Given:

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

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

$\Rightarrow$$\sin \left(\theta \right)[1+{\sin }^2(\theta \left)\right]={\cos }^2\left(\theta \right)$

$\Rightarrow$$\sin \left(\theta \right)[2-{\cos }^2(\theta \left)\right]={\cos }^2\left(\theta \right)$

Squaring both sides,

$\Rightarrow$${\sin }^2\left(\theta \right)[4+{\cos }^4(\theta )-4{\cos }^2(\theta \left)\right]={\cos }^4\left(\theta \right)$

$\Rightarrow$$[1-{\cos }^2(\theta \left)\right][4+{\cos }^4(\theta )-4{\cos }^2(\theta \left)\right]={\cos }^4\left(\theta \right)$

$\Rightarrow$$4+{\cos }^4\left(\theta \right)-4{\cos }^2\left(\theta \right)-4{\cos }^2\left(\theta \right)-{\cos }^6\left(\theta \right)+4{\cos }^4\left(\theta \right)={\cos }^4\left(\theta \right)$

Rearranging the terms, we get

$8{\cos }^2\left(\theta \right)-4{\cos }^4\left(\theta \right)+{\cos }^6\left(\theta \right)=4$