$\begin{aligned} & \text {} f_n(x)=\frac{1}{2 n}\left[\sin ^{2 n} x+\cos ^{2 n} x\right] \\ & f_1(x)+f_2(x)+f_3(x) \\ & =\frac{1}{2}\left[\sin ^2 x+\cos ^2 x\right]+\frac{1}{4}\left[\sin ^4 x+\cos ^4 x\right]-\frac{1}{6}\left[\sin ^6 x+\cos ^6 x\right]
\\ & =\frac{1}{2} \times 1+\frac{1}{4}\left[\sin ^4 x+\cos ^4 x\right]-\frac{1}{6}\left[\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^4 x+\cos ^4 x-\sin ^2 x \cos ^2 x\right)\right]
\\ & =\frac{1}{2}+\frac{1}{4}\left[\sin ^4 x+\cos ^4 x\right]-\frac{1}{6}\left[\sin ^4 x+\cos ^4 x-\sin ^2 x \cos ^2 x\right]
\\ & =\frac{1}{2}+\frac{1}{12}\left[\sin ^4+\cos ^4 x\right]+\frac{1}{6} \sin ^2 x \cos ^2 x \\ & =\frac{1}{2}+\frac{1}{12}\left[\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x\right]+\frac{1}{6}\sin ^2 x \cos ^2 x
\\ & =\frac{1}{2}+\frac{1}{12} \times 1-\frac{1}{6} \sin ^2 x \cos ^2 x+\frac{1}{6} \sin ^2 x \cos ^2 x \\ & =\frac{1}{2}+\frac{1}{12}=\frac{6+1}{12}=\frac{7}{12}\end{aligned}$