Let f_ k (x)=(1)/(k) ( ^ k x+ ^ k x) for k=1,2,3, Then for all x ∈...

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