Question

If $\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}$, then show that $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}=\frac{1}{625}$.

Answer 1

$\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}=>\frac{{\sin }^{4}x}{2}+\frac{{(1-{\sin }^{2}x)}^2}{2}=\frac{1}{5}let\sin x=t\therefore \frac{t^4}{2}+\frac{{(1-t^2)}^2}{3}=\frac{1}{5}\therefore \frac{3t^4+2(1-{2t}^2+t^4)}{6}=\frac{1}{5}=>15t^4+10+10t^4-20t^2-6=0=>25t^4-20t^2+4=0=>(5t^2-2{)}^2=0=>5t^2=2=>t^2=\frac{2}{5}=>1-{\cos }^{2}x=\frac{2}{5}=>{\cos }^{2}x=\frac{3}{5}ThusLHS=\frac{{\left({\sin }^{2}x\right)}^4}{8}+\frac{{\left({\cos }^{2}x\right)}^4}{27}=>\frac{2^4}{5^4\times 8}+\frac{3^4}{5^4\times 27}=>\frac{2}{625}+\frac{3}{625}=>\frac{1}{625}=RHS$

Hence $LHS=RHS$