I was trying to solve a complex number to a power using de Moivre's Identity but got stuck. One of the assignments that I have to solve is:
$(7-i)^3$
I know that I can simply open the exponent since it is only 3 but I tried to use de Moivre's Identity.
$\theta=\tan^{-1}(\frac{-1}{7})$
$r=\sqrt{7^{2}+(-1)^{2}}=\sqrt{50}$
$z^{n}=r(\cos\theta+i\sin\theta)=50^{\frac{3}{2}}(\cos(3\cdot\tan^{-1}(\frac{-1}{7}))+i\sin(3\cdot\tan^{-1}(\frac{-1}{7})))$
How can I find out what the cosine of an arc-tangent is?

Well, we have $\arctan(-x)=-\arctan(x)$, so $\arctan\left(\dfrac{-1}{7}\right)=-\arctan\left(\dfrac{1}{7}\right)$ and we also can simplify $\sqrt{50}=\sqrt{2\times25}=5\sqrt{2}$. Then your last expression simplifies to $$(7-i)^3=5\sqrt{2}\cos\left(3\arctan\left(\frac{1}{7}\right)\right)-5\sqrt{2}i\sin\left(3\arctan\left(\frac{1}{7}\right)\right)$$
Now, to really answer your question, first we need to get rid of that $3$ factor using the following trigonometric identities: $$\sin(3x)=3\sin(x)\cos(x)^2-\sin(x)^3,$$$$\cos(3x)=\cos(x)^3-3\cos(x)\sin(x)^2;$$ these can easily be derived for arbitrary integer factor once you learn about complex exponentiation and generally for arbitrary (symbolic) factor using the binomial theorem.
We have $$(7-i)^3=5\sqrt{2}\left(\cos\left(\arctan\left(\frac{1}{7}\right)\right)^3-3\cos\left(\arctan\left(\frac{1}{7}\right)\right)\sin\left(\arctan\left(\frac{1}{7}\right)\right)^2\right)-5\sqrt{2}i\left(3\sin\left(\arctan\left(\frac{1}{7}\right)\right)\cos\left(\arctan\left(\frac{1}{7}\right)\right)^2-\sin\left(\arctan\left(\frac{1}{7}\right)\right)^3\right);\label{eq1}\tag{1}$$
now we can evaluate $\sin\circ\arctan$ and $\cos\circ\arctan$: just recall
$$\sin(x)^2+\cos(x)^2=1;$$
we want to have $\sin$ and $\cos$ in terms of $\tan$ to cancel $\arctan$, therefore divide by $\sin(x)^2$:
$$1+1/\tan(x)^2=1/\sin(x)^2\implies$$
$$\implies \sin(x)^2=\frac{1}{1+1/\tan(x)^2}=\frac{\tan(x)^2}{\tan(x)^2+1}$$
$$\implies \sin(x)=\pm\frac{\tan(x)}{\sqrt{\tan(x)^2+1}},$$
$$\sin(\arctan(\alpha))=\frac{\tan(\arctan(\alpha))}{\sqrt{\tan(\arctan(\alpha))^2+1}}=\frac{\alpha}{\sqrt{\alpha^2+1}};$$
and $\cos(x)^2$:
$$\tan(x)^2+1=1/\cos(x)^2\implies$$
$$\implies\cos(x)^2=\frac{1}{\tan(x)^2+1}$$
$$\implies\cos(x)=\pm\frac{1}{\tan(x)^2+1},$$
$$\cos(\arctan(\alpha))=\frac{1}{\sqrt{\tan(\arctan(\alpha))^2+1}}=\frac{1}{\sqrt{\alpha^2+1}};$$
in particular then, we have
$$\sin\left(\arctan\left(\frac{1}{7}\right)\right)=\frac{1/7}{\sqrt{1/7^2+1}}=\frac{1}{7\sqrt{1/7^2+1}}=\frac{1}{\sqrt{7^2+1}}=50^{-1/2}=\frac{1}{5}2^{-1/2},$$
$$\cos\left(\arctan\left(\frac{1}{7}\right)\right)=\frac{1}{\sqrt{1/7^2+1}}=\frac{7}{\sqrt{7^2+1}}=\frac{7}{5}2^{-1/2};$$
substituting that in $\eqref{eq1}$ will yield the desired form after a bit of simplification.
