Complex numbers and trig identities: $\cos(3\theta) + i \sin(3\theta)$ Using the equally rule $a + bi = c + di$ and trigonometric identities how do I make...
$$\cos^3(\theta) - 3\sin^2(\theta)\ \cos(\theta) + 3i\ \sin(\theta)\ \cos^2(\theta) - i\ \sin^3(\theta)=
\cos(3\theta) + i\ \sin(3\theta)$$
Apparently it's easy but I can't see what trig identities to substitute
PLEASE HELP!
 A: Note that $$(\cos(t)+i\sin(t))^n=(\cos(nt)+i\sin(nt)),~~n\in\mathbb Z$$ and $(a+b)^3=a^3+3a^2b+3ab^2+b^3,~~~(a-b)^3=a^3-3a^2b+3ab^2-b^3$.
A: Use the fact that
$$\cos{(n x)} + i \sin{( n x)} = (\cos{x} + i \sin{x})^n$$
You are interested in the case $n=3$.  Expand the binomial and equate real and imaginary parts.
A: From the wording, it appears that you are asked to solve the problem without using DeMoivre's Theorem!
Then you need information about $\cos 3x$ and $\sin 3x$ from another source. Luckily, you probably have all the tools needed. 
For $\cos 3x$, write it as $\cos(2x+x)$, and use the usual cosine of a sum rule. We get $\cos 2x\cos x-\sin 2x\sin x$. Using the double-angle identities, you then get $(\cos^2 x-\sin^2 x)\cos x-2\sin^2 x\cos x$.   I used the rule $\cos 2x=\cos^2 x-\sin^2 x$ because it fits in well with what you are asked to prove. 
We conclude that $\cos 3x=\cos^3 x-3\sin^2 x\cos x$.
You will need similar information about $\sin 3x$. Be guided by the ultimate result you are looking for. But start with $\sin 3x=\sin(2x+x)=\sin 2x \cos x+\cos 2x\sin x$. Of course you will then use $\sin 2x=2\sin x\cos x$, and an appropriate identity for $\cos 2x$.
A: $(\cos x+i\sin x)^3=(\cos 3x+i\sin 3x)$(by De Moivre's theorem)
But $(\cos x+i\sin x)^3=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i\cos^2 x\sin x-3\cos x\sin^2x=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i(1-\sin^2x) \sin x-3\cos x(1-\cos^2 x)=4\cos ^3x-3\cos x+i(3\sin x-4\sin^3x)$
Now by equating we have, $\cos 3x=4\cos ^3x-3\cos x$ and $\sin 3x=3\sin x-4\sin^3x$
