Question

If $\sin A+{\sin }^{2}A+{\sin }^{3}A=1$, then find the value of ${\cos }^{6}A-4{\cos }^{4}A+8{\cos }^{2}A$.

Answer 1

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

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

$\sin A(1+{\sin }^{2}A)={\cos }^{2}A$

$\sin A(2-{\cos }^{2}A)={\cos }^{2}A$ (since, ${\sin }^{2}A=1-{\cos }^{2}A$)

Squaring both sides,

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

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

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

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

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