Complex Analysis: why does $\cos(3\theta)$ = $\cos^3\theta - 3\cos\theta \sin^2\theta$. The formulas are stated in my text book but they are not proven.  There is nothing in the preceding paragraphs that offers any insight other than "this could be proven using basic properties of complex numbers".  
I know that complex numbers can be expressed in terms of $\sin$ and $\cos$ but usually its both.  If I have just $\cos3 \theta$ , I'm really not sure how to go about starting.  
 A: $$e^{3it}=\left(e^{it}\right)^3\iff \cos3t+i\sin3t=\left(\cos t+i\sin t\right)^3\iff$$
$$\cos3t+i\sin3t=\cos^3t+3i\cos^2t\sin t-3\cos x\sin^2t-i\sin^3t$$
and now just compare real parts in both sides.
A: The sophisticated proof is to expand $(\cos\theta+i\sin\theta)^3$ by the binomial theorem. The unsophisticated proof is repeated use of $\cos(A+B)=\cos A\cos B-\sin A\sin B$ and $\sin(A+B)=\sin A\cos B+\sin B\cos A$.
The first works because $\cos\theta+i\sin\theta=e^{i\theta}$.
A: We have $$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$$ and $$sin(x+y)=sin(x)cos(y)+sin(y)cos(x)$$ for all $x,y\in \mathbb C$
It follows
$$cos(2x)=cos^2x-sin^2x$$
Now, we have $$cos(3x)=cos(x)cos(2x)-sin(x)sin(2x)=cos^3x-sin^2xcos(x)-sin(x)\cdot 2sin(x)cos(x)=cos^3x-3sin^2xcos(x)$$
A: By Euler's relation
$$ e^ { i\, 3  \theta  } = \cos 3 \theta + i \sin 3 \theta $$
the LHS can be also expressed as
$$ (\cos  \theta  + i \sin \theta)^3  = (\cos  3\theta  + i \sin 3\theta) $$
Expand LHS as a cubic term by term:
$$ =\cos ^3 \theta  + 3  \cos ^2 \theta  \cdot i \sin \theta + 3  \cos  \theta \cdot i^2   \sin ^2 \theta   + i^3 \sin ^3 \theta  $$
To find $ \cos 3 \theta  $ equate real parts (first and third terms ) and to find $ \sin 3 \theta  $ equate imaginary parts (second and fourth terms). We can recognize such derivational origin by polynomial approach right away. 
A: Use de Moivre's formula on $e^{3it}$:
$$\left(e^{it}\right)^3 = e^{3it} = e^{i(3t)}$$
Then:
$$e^{3it}=\left(e^{it}\right)^3=\left(\cos t+i\sin t\right)^3 \tag{1}$$
$$e^{3it} = e^{i(3t)} = \cos3t+i\sin3t \tag{2}$$
By expanding (1), we get:
$$(1) = \cos^3t+3i\cos^2t\sin t-3\cos x\sin^2t-i\sin^3t$$
Now compare $(1)$ and $(2)$ deduce not only a formula for $\cos(3t)$ but also a formula for $\sin(3t)$.
Also, check out this old revision of a question: https://math.stackexchange.com/revisions/2866072/2
