\tan^4 2x &amp;= \frac{\sin^4 2x}{\cos^4 2x} \\&amp;=\frac{\frac{1}{8}\cos 8x - \frac{1}{2}\cos 4x + \frac{3}{8}}{\frac{1}{8}\cos 8x+\frac{1}{2}\cos 4x+\frac{3}{8}} \\&amp;=\frac{\cos 8x - 4\cos 4x + 3}{\cos 8x+4\cos 4x+3}