Question

Show that :
${\cos }^{6}A-4{\cos }^{4}A+8{\cos }^{2}A=4$

Answer 1

If $\sin A+{\sin }^{2}A+{\sin }^{3}A=1$ then$,$

$\sin A+{\sin }^{3}A=1-{\sin }^{2}A$ $={\cos }^{2}A$

$\sin A\left(2-{\cos }^{2}A\right)$

${\sin }^{2}A\left(4+{\cos }^{4}A-4{\cos }^{2}A\right)={\cos }^{4}A$

$\left(1-{\cos }^{2}A\right)\left(4-4{\cos }^{4}A-4{\cos }^{2}A\right)={\cos }^{4}A$

$4-4{\cos }^{2}A+{\cos }^{2}A-4{\cos }^{2}A-{\cos }^{6}A+4{\cos }^{4}A={\cos }^{4}A$

$4-{\cos }^{6}A+4{\cos }^{4}A-8{\cos }^{2}A=0$

${\cos }^{6}A-4{\cos }^{4}A+8{\cos }^{2}A=4$ Proved$.$