Identities for $\sin(2x)$ and $\sin(3x)$, as well as their cosine counterparts are very common, and can be used to synthesize identities for $\sin(4x)$ and above. Given some integer $k$, is there an equation to find $\sin(kx)$ in terms of $\sin(x)$ and $\cos(x)$?
For example, if I wanted to find the identity for $\sin(1000x)$, what would it be?
Identies used: $$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\\\cos(a+b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$
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There's no neat formula. However, $\sin( k\theta)=\sin\theta\cdot \ U_{k-1}(\cos\theta)$ where $U_{k-1}(x)$ is a polynomial of degree $k-1$. This polynomial $U_{k-1}(x)$ is also known as the Chebyshev polynomial of the second kind having degree $k-1$.
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Hint. One has $$ \sin n \theta= U_{n-1}(\cos \theta) \cdot \sin \theta $$ with $$ U_n(x)=\sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} (-1)^k \binom{n-k}{k}~(2x)^{n-2k} $$ these polynomials are known as Chebyshev polynomials of the second kind.
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$$ \sin nx={\rm Im\;} e^{i n x}={\rm Im\;} \left(e^{ix}\right)^n={\rm Im\;} \left(\cos x+i\sin x\right)^n={\rm Im\;}\sum_{k=0}^n\binom{n}{k}\cos^{n-k} x(i\sin x)^{k}\\=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^k\binom{n}{2k+1}\cos^{n-2k-1} x\sin^{2k+1} x. $$ Similarly $$ \cos nx=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k} x\sin^{2k} x. $$
Use the De Moivre formula to calculate $\sin(nx)$ and $\cos(nx)$ for large $n$