If $n$ is an odd natural number, and $\sin(n\theta) = \Sigma_{r=0}^{n} b_r \sin^r\theta$, then find $b_r$ in terms of $n$.
I have tried this using trigonometric expansion but unable to find solution for this. Can body help me to solve this trigonometric problem?
On
Hints: 1) $\forall x\in\mathbb{R},\,\sin (x)=\dfrac{e^{ix}-e^{-ix}}{2i}$ (Euler's Formula).
2) $\forall x\in\mathbb{R},\,\forall n\in\mathbb{N},\,(\cos x+i\sin x)^n=\cos nx+i\sin nx$ (Moivre's Formula). So $\sin nx=\Im\left((\cos x+i\sin x)^n\right)$ where for $z\in\mathbb{C}$, $\Im(z)$ is the imaginary part of $z$.
3) $\forall a,b\in\mathbb{C},\,\forall n\in\mathbb{N},\,(a+b)^n=\sum\limits_{k=0}^n\binom{n}{k}a^kb^{n-k}$ (Binomial Theorem).
4) From Euler's formula we have $\sin nx=\dfrac{e^{inx}-e^{-inx}} {2}=\dfrac{(e^{ix})^n-(e^{-ix})^n}{2}$ then use Moivre's Formula and the Binomial Theorem to get a sum where you have powers of $\cos$ and $\sin$ and something in term of $i$. Recall that $i=e^{i\frac{\pi}{2}}$. The final formula you should get is:$$\sin n\theta=\sum_{k=0}^n\binom{n}{k}\cos^k \theta\sin ^{n-k}\theta\sin\left( \dfrac{n-k}{2}\pi\right)$$
This formula was given by Franciscus Vieta.
A Chebyshev polynomial of the first kind is defined as $$ T_n(\cos \phi) = \cos n \phi $$ Lets take $\phi = \frac{\pi}{2} - \theta$. Then $\cos \phi = \sin \theta$ and $$ T_{2k+1}(\sin \theta) = \cos (2k+1) \left(\frac{\pi}{2} - \theta\right) = \cos \left(\frac{(2k+1)\pi}{2} - (2k+1) \theta\right) = \sin \left((2k+1) \theta - k\pi\right) = \\ =(-1)^k \sin (2k+1) \theta. $$ Finally, $$ \sin (2k+1) \theta = (-1)^k T_{2k+1}(\sin \theta) $$
PS. Basically, I took expansion of $\sin 3x, \sin 5x, \sin 7x$ and searched for the coefficients in OEIS and found the sequence A084930.