Showing Trigonometric Identity Prove that:
$$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$
Attempt:
\begin{align*}
L.H.S & = \cos^2\theta\sin^4\theta\\
& = \cos^2\theta\sin^2\theta\sin^2\theta\\
& = \frac{1+\cos2\theta}{2}.\frac{1-\cos2\theta}{2}.\frac{1-\cos2\theta}{2}\\
& = \frac{1}{8} (1-\cos^22\theta)(1-\cos2\theta)
\end{align*}
Now, what should I do?
 A: 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*}
A: HINT: use that $$\cos(6\theta)=32\, \left( \cos \left( \theta \right)  \right) ^{6}-48\, \left( \cos
 \left( \theta \right)  \right) ^{4}+18\, \left( \cos \left( \theta
 \right)  \right) ^{2}-1
$$
$$\cos(2\theta)=2\, \left( \cos \left( \theta \right)  \right) ^{2}-1$$
$$\cos(4\theta)=8\, \left( \cos \left( \theta \right)  \right) ^{4}-8\, \left( \cos
 \left( \theta \right)  \right) ^{2}+1
$$
A: With $z=e^{i\theta}$,
$$\left(\frac{z+z^{-1}}2\right)^2\left(\frac{z-z^{-1}}{2i}\right)^4=\frac1{32}\left(\frac{z^6+z^{-6}-z^2-z^{-2}+4-2z^4-2z^{-4}}2\right).$$
For convenience, multiply by $64z^6$, set $t:=z^2$ and rewrite
$$(t^2-1)^2(t-1)^2=t^6+1-t^4-t^2+4t^3-2t^5-2t.$$
By direct expansion of the LHS you get the equality.
A: 
$$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$

Use $\sin^2 \theta + \cos^2 \theta = 1$ to put everything in terms of cosine:
$$\begin{align}
\cos^2\theta (1 - \cos^2 \theta )^2               &= \frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta) \\
\cos^2\theta (1 - 2\cos^2\theta + \cos^4 \theta ) &= \frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta) \\
\cos^2\theta - 2\cos^4\theta + \cos^6 \theta      &= \frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)
\end{align}$$
Use $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ :
$$
  \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^2
- 2\left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^4
+ \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^6
= 
\frac{1}{32}\left(
\left(\frac{e^{6i\theta} + e^{-6i\theta}}{2}\right) - 
\left(\frac{e^{2i\theta} + e^{-2i\theta}}{2}\right)
+ 2
- 2\left(\frac{e^{4i\theta} + e^{-4i\theta}}{2}\right)
\right)$$
Multiply it out:
$$
\begin{align} 
  \frac{1}{4}  & \left( e^{2i\theta} + 2 + e^{-2i\theta} \right)  \\
- \frac{1}{8}  & \left( e^{4i\theta} + 4 e^{2i\theta} + 6 + 4 e^{-2i \theta} + e^{-4i\theta} \right) \\
+ \frac{1}{64} & \left( e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \right) \\
= 
\frac{1}{32} & \left(
\frac{
  \left( e^{6i\theta} + e^{-6i\theta} \right) - 
  \left( e^{2i\theta} + e^{-2i\theta} \right)
  + 4
  - \left(2e^{4i\theta} + 2e^{-4i\theta}\right)
}{2}
\right)
\end{align}$$
Combine like terms:
$$\begin{array} {c}
  16  e^{2i\theta} + 32 + 16 e^{-2i\theta} \\
-~ 8  e^{4i\theta} - 32 e^{2i\theta} - 48 - 32 e^{-2i \theta} - 8 e^{-4i\theta} \\
+~ e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \\
=~ 
     e^{6i\theta}
  - 2e^{4i\theta}
  - e^{2i\theta}
  + 4
  - e^{-2i\theta}
  - 2e^{-4i\theta}
   + e^{-6i\theta} 
\end{array}$$
So
$$\begin{array} {c}
e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta} \\
= \\
e^{6i\theta} + 6 e^{4i\theta} + 15 e^{2i\theta} + 20 + 15 e^{-2i\theta} + 6 e^{-4i\theta} + e^{6i\theta}
\end{array}$$
