Solution
[tex]\begin{gathered} \tan^2(4x)=\frac{1-\cos(8x)}{1+\cos(8x)} \\ \\ \Rightarrow\tan^4(4x)=\frac{1-2\cos(8x)+\cos^2(8x)}{1+2\cos(8x)+\cos^2(8x)} \\ \\ \text{ since }\cos^2(8x)=\frac{1+\cos(16x)}{2} \\ \\ \Rightarrow\tan^4(4x)=\frac{1-2\cos(8x)+\frac{1+\cos(16x)}{2}}{1+2\cos(8x)+\frac{1+\cos(16x)}{2}} \\ \\ \Rightarrow\tan^4(4x)=\frac{2-4\cos(8x)+1+\cos(16x)}{2+4\cos(8x)+1+\cos(16x)} \\ \\ \Rightarrow\tan^4(4x)=\frac{3-4\cos(8x)+\cos(16x)}{3+4\cos(8x)+\cos(16x)} \end{gathered}[/tex]
The answer is:
[tex]\frac{3-4\cos(8x)+\cos(16x)}{3+4\cos(8x)+\cos(16x)}[/tex]
