Question

If $\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}$ then $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}$

Answer 1

Given$:\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}$

$\Rightarrow \frac{{\sin }^{4}x}{2}+\frac{{(1-{\sin }^{2}x)}^2}{3}=\frac{1}{5}$

Let $\sin x=t$

$\Rightarrow \frac{t^4}{2}+\frac{{(1-t^2)}^2}{3}=\frac{1}{5}$

$\Rightarrow \frac{3t^4+2{(1-t^2)}^2}{6}=\frac{1}{5}$

$\Rightarrow \frac{3t^4+2(1+t^4-2t^2)}{6}=\frac{1}{5}$

$\Rightarrow 15t^4+10{(1+t^4-2t^2)}^2-6=0$

$\Rightarrow 25t^4-20t^2+4=0$

$\Rightarrow {\left(5t^2\right)}^2-2\times 5t^2\times 2+4=0$

$\Rightarrow {(5t^2-2)}^2=0$

$\Rightarrow 5t^2-2=0$

$\Rightarrow t^2=\frac{2}{5}$

$\Rightarrow {\sin }^{2}x=\frac{2}{5}$

$\Rightarrow {\cos }^{2}x=1-{\sin }^{2}x=1-\frac{2}{5}=\frac{5-2}{5}=\frac{3}{5}$

$\Rightarrow \frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{{\left({\sin }^{2}x\right)}^4}{8}+\frac{{\left({\cos }^{2}x\right)}^4}{27}$

$=\frac{{\left(\frac{2}{5}\right)}^4}{8}+\frac{{\left(\frac{3}{5}\right)}^4}{27}$

$=\frac{16}{625\times 8}+\frac{81}{625\times 27}$

$=\frac{1}{625}(2+3)=\frac{5}{625}=\frac{1}{125}$