Trigonometric Identities for $\sin nx$ and $\cos nx $ These are generalizations of simple trigonometric identities for $\sin 2x$ and $\cos 2x$, but in general how can we prove them? 
$$\sin nx =\sum_{k=1}^{\left\lceil\frac{n}{2}\right\rceil}(-1)^{k-1}\binom{n}{2k-1}\sin^{2k-1}x\cos^{n-2k+1}x,$$
$$\cos nx =\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}(-1)^{k}\binom{n}{2k}\sin^{2k}x\cos^{n-2k}x,$$
$$\sin nx =\sin x\left(\sum_{k=1}^{\left\lceil\frac n 2\right\rceil}(-1)^{k-1}\binom{n-k}{k-1}(2\cos x)^{n-2k+1}\right),$$
$$\cos nx =\cos x\left((2\cos x)^{n-1}+\sum_{k=1}^{\left\lfloor\frac n 2\right\rfloor}\frac n k(-1)^k\binom{n-k-1}{k-1}(2\cos x)^{n-2k-1}\right).$$

Have a look at http://en.wikipedia.org/wiki/Chebyshev_polynomials. Once you get the first two identities, you'll easily get the last two. As pbs suggested, you can use the Binomial Theorem. The reason is :
$$\cos(n\theta) + i\sin(n\theta) = e^{in\theta} = (e^{i\theta})^{n} = (\cos(\theta)+i\sin(\theta))^{n}$$
The idea is to expand $(\cos(\theta) + i\sin(\theta))^{n}$ to identify its real and imaginary part. In short, you have :
\begin{eqnarray*}
(\cos(\theta)+i\sin(\theta))^{n} & = & \sum_{k=0}^{n} \begin{pmatrix} n \\ k \end{pmatrix} i^{k} \sin^{k}(\theta) \cos^{n-k}(\theta) \\
 & = & \sum_{p=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^{p} \begin{pmatrix} n \\ 2p \end{pmatrix} \sin^{2p}(\theta)\cos^{n-2p}(\theta) - i\sum_{p=0}^{\lfloor \frac{n+1}{2} \rfloor} (-1)^{p} \begin{pmatrix} n \\ 2p-1 \end{pmatrix} \sin^{2p-1}(\theta)\cos^{n-2p+1}(\theta)
\end{eqnarray*}
So, you get the expected identities :
$$\cos(n\theta) = \sum_{p=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^{p} \begin{pmatrix} n \\ 2p \end{pmatrix} \sin^{2p}(\theta)\cos^{n-2p}(\theta)$$
and 
$$ \sin(n\theta) = \sum_{p=0}^{\lceil \frac{n}{2} \rceil} (-1)^{p+1} \begin{pmatrix} n \\ 2p-1 \end{pmatrix} \sin^{2p-1}(\theta)\cos^{n-2p+1}(\theta)$$
