Express $\sin nx$ and $\cos nx$ in terms of $\sin x$ and $\cos x$ respectively 
What are the expansions of $\sin nx $ and $\cos nx$ in terms of $\sin x$ and $\cos x$ respectively? (here $n \in \mathbb N$).

Maybe  this  is  solved  problem  or  there  is  new  technique  to  answer  this  question
Tanhks
 A: Let $T_n$ be the $n$th Chebychev polynomial, then
$$T_n(\cos(x))=\cos(nx)$$
For more information see here. The explicit formulas you can find here.
A: $$
\cos(n x) =\frac{1}{2} \sum _{k=0}^{\infty } i^{-k-n} \cos ^k(z) \sin ^{-k-n}(z) \left[\binom{-n}{k}+i^{2 n} \binom{n}{k} \sin ^{2 n}(z)\right]
$$
$$
\sin(n x) = \frac{1}{2 i}\sum _{k=0}^{\infty } i^{-k-n} \cos ^k(z) \sin ^{-k-n}(z) \left[-\binom{-n}{k}+i^{2 n} \binom{n}{k} \sin ^{2 n}(z)\right]
$$
and
$$
\cos(n x) =\frac{1}{2} \sum _{k=0}^{\infty } i^{-k-n} \cos ^k(z)  (1-\cos^2(x))^{(-k-n)/2} \left[\binom{-n}{k}+i^{2 n} \binom{n}{k} (1-\cos^2(x)) ^{ n}(z)\right]
$$
$$
\sin(n x) = \frac{1}{2 i}\sum _{k=0}^{\infty } i^{-k-n} (1-\sin^2(z)) ^{k/2} \sin ^{-k-n}(z) \left[-\binom{-n}{k}+i^{2 n} \binom{n}{k} \sin ^{2 n}(z)\right]
$$
A: Note De'Moivre's formula:$$\cos(n x)+i\sin(n x) = (\cos(x)+i\sin(x))^n.$$ You can use the Binomial Theorem in the right to explore further and take either real or imaginary parts to isolate for cosine and sine as you require. See here and here for further details.
EDIT
After the expansion you can substitute $\cos(x)=\sqrt{1-\sin^2(x)}$ to obtain an expression purely in terms of sines.
