Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

$f_k(x)=\frac{1}{k}\left({\sin }^{k}x+{\cos }^{k}x\right)$ $f_4(x)=\frac{1}{4}\left[{\sin }^{4}x+{\cos }^{4}x\right]$ $=\frac{1}{4}\left[{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-\frac{(\sin 2x{)}^2}{2}\right]$ $=\frac{1}{4}\left[1-\frac{(\sin 2x{)}^2}{2}\right]$ $f_6(x)=\frac{1}{6}\left[{\sin }^{6}x+{\cos }^{6}x\right]$ $=\frac{1}{6}[\left({\sin }^{2}x+\left({\cos }^{2}x\right)-\frac{3}{4}{\left({\sin }^{2}x\right)}^2\right]$ $=\frac{1}{6}\left[1-\frac{3}{4}(\sin 2x{)}^2\right]$ Now $f_4(x)-f_{(6)}(x)=\frac{1}{4}-\frac{1}{6}-\frac{(\sin 2x{)}^2}{8}+\frac{1}{8}(\sin 2x{)}^2$ $=\frac{1}{12}$