Need help using De Moivre's theorem to write $\cos 4\theta$ & $\sin 4\theta$ as terms of $\sin\theta$ and $\cos\theta$ I need help with the following question: 
"Use De Moivre's theorem to write $\cos 4\theta$ & $\sin 4\theta$ as terms of $\sin\theta$ and $\cos\theta$"
You could write the problem as:
$(\cos\theta+i\sin\theta)^4 - i\sin 4\theta$ = $\cos 4\theta$
I don't think it is ment that you should develop the $exp 4$ parenthesis, so there must be a simpler way? What would be your strategy with approaching this problem?
Thank you
 A: Use the binomial theorem for $(x+i y)^4$:
$$\begin{align}(x+i y)^4 &= x^4 + 4 i x^3 y +6 i^2 x^2 y^2+4 i^3 x y^3 + i^4 y^4\\&=  x^4-6 x^2 y^2 + y^4 + i (4 x^3 y - 4 x y^3) \end{align}$$
using $i^2=-1$, etc.  Now let $x=\cos{\theta}$, $y=\sin{\theta}$:
$$e^{i 4 \theta} = \cos{4 \theta} + i \sin{4 \theta} =  (\cos{\theta} + i \sin{\theta})^4 $$
so that
$$\cos{4 \theta} = \cos^4{\theta} - 6 \cos^2{\theta} \sin^2{\theta} + \sin^4{\theta}$$
$$\sin{4 \theta} = 4 \cos{\theta} \sin{\theta} (\cos^2{\theta} - \sin^2{\theta}) = 2 \sin{2 \theta} \cos{2 \theta}$$
Note also that $\cos{4 \theta}$ could be derived from
$$\cos{4 \theta} = \cos^2{2 \theta} - \sin^2{2 \theta}$$
A: $$\cos(4\theta)+i\sin(4\theta)=(\cos(\theta)+i\sin(\theta))^4=\sum_{k=0}^4\binom{4}{k}(\cos(\theta))^k(i\sin(\theta))^{4-k}$$
$$\cos(4\theta)=Re(\cos(4\theta)+i\sin(4\theta))=Re(\sum_{k=0}^4\binom{4}{k}(\cos(\theta))^k(i\sin(\theta))^{4-k})$$
Thus:
$$\cos(4\theta)=Re(\sum_{k=0}^4\binom{4}{k}(\cos(\theta))^k(i\sin(\theta))^{4-k})=\cos^4(\theta)-6\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)$$
To get $\sin(4\theta)$, compare the imaginary parts
