Showing Trigonometric Identity

Use the linearisation and the duplication formulae: \begin{align*} \cos^2\theta\sin^4\theta & = \cos^2\theta\sin^2\theta\sin^2\theta = \frac14 \sin^2 2\theta\sin^2\theta\\ &=\frac14\frac{1-\cos4\theta}{2}\frac{1-\cos2\theta}{2}= \frac{1}{32} (2\cos4\theta\cos2\theta-2\cos4\theta-2\cos2\theta+2)\\ &=\frac1{32}(\cos 6\theta+\cos2\theta-2\cos4\theta-2\cos2\theta+2)\\ &=\frac1{32}(\cos 6\theta-2\cos4\theta- \cos2\theta+2)\ \end{align*}

If you are allowed to use complex numbers, this is much easier: set $u=\mathrm e^{\mathrm i\theta}$. Then we have, by Euler's formulae: $$\bar u=\mathrm e^{-\mathrm i\theta},\quad \cos\theta=\frac{u+\bar u}2,\quad \sin\theta =\frac{u-\bar u}{2\mathrm i},$$ whence (note $u\bar u=1$) \begin{align*} \cos^2\theta\sin^4\theta & = \frac{(u+\bar u)^2}4 \frac{(u-\bar u)^4}{16}=\frac1{64}(u^2-\bar u^2)^2(u-\bar u)^2\\ &=\frac1{64}(u^4-2+\bar u^4)(u^2-2+\bar u^2)\\ &=\frac1{64}(u^6-2u^4+u^2-2u^2+4-2\bar u^2+\bar u^2-2\bar u^4+\bar u^6)\\ &=\frac1{64}(u^6+\bar u^6-2(u^4+\bar u^4) -(u^2+\bar u^2)+4)\\ &=\frac1{32}(\cos6\theta-2\cos4\theta-\cos2\theta+2). \end{align*}