Proof that $\sin^{2n + 1}(x) = \sum_{i = 0}^{n} a_{i} \sin\left(\left(2i + 1\right)x\right)$ In playing around on wolframalpha, I noticed that $\sin^{2n + 1}(x)$ where $n \in \mathbb{N} \cup \{0\}$ takes the form:
$$\sin^{2n + 1}(x) = \sum_{i = 0}^{n} a_{i} \sin\left(\left(2i + 1\right)x\right)$$
Where $a_{i} \in \mathbb{Q}$ 
Does anyone have any ideas on how one would go about (dis)proving this?
Any tips would be greatly appreciated. 
 A: Just a thought that came to mine on how this could be proven
Employ proof my induction 
Step (1) : Prove true for $n = 0 $
$$\sin^{1}(x) = 1\cdot\sin(1\cdot x) = \sum_{i = 0}^{0} a_{i}^{(0)}\sin\left(\left(2i + 1\right)x\right)$$
Hence true. 
Step (2) : Inductive Step - Assume true for $n = k$:
$$\sin^{2k + 1}(x) = \sum_{i = 1}^{k + 1}a_{i}^{(k)}\sin\left(\left(2i + 1\right)x\right)$$
Step (3) : Prove true for $n = k + 1$. Hence show, 
$$\sin^{2(k + 1) + 1}(x) = \sum_{i = 1}^{k + 1}a_{i}^{(k + 1)}\sin\left(\left(2i + 1\right)x\right)$$
Now, 
\begin{align}
\sin^{2k + 3}(x) &= \sin^{2}\left(x\right)\sin^{2k + 1}(x)  \\
&= \left[\frac{1 - \cos(2x)}{2} \right]\sum_{i = 0}^{k}a_{i}^{(k)}\sin\left(\left(2i + 1\right)x\right) \\
&= \sum_{i = 0}^{k}\frac{a_{i}^{(k)}}{2}\sin\left(\left(2i + 1\right)x\right) - \sum_{i = 0}^{k }\frac{a_{i}^{(k)}}{2}\cos(2x)\sin\left(\left(2i + 1\right)x\right) \\
&= \sum_{i = 0}^{k}\frac{a_{i}^{(k)}}{2}\sin\left(\left(2i + 1\right)x\right) - \sum_{i = 0}^{k}\frac{a_{i}^{(k)}}{4}\left[\sin\left(\left(2i + 3\right)x\right) + \sin\left(\left(2i - 1\right)x\right) \right]
 \\
&= \sum_{i = 0}^{k}\frac{a_{i}^{(k)}}{2}\sin\left(\left(2i + 1\right)x\right) - \sum_{i = 1}^{k + 1}\frac{a_{i-1}^{(k)}}{4}\sin\left(\left(2i + 1\right)x\right) -\sum_{i = -1}^{k - 1}\frac{a_{i + 1}^{(k)}}{4} \sin\left(\left(2i + 1\right)x\right) \\
&=\sum_{i = 1}^{k - 1}\left[\frac{2a_{i}^{(k)} - a_{i - 1}^{(k)} - a_{i+1}^{(k)}}{4}\right]\sin\left(\left(2i + 1\right)x\right) + \left[\frac{3a_{0}^{(k)} - a_{1}^{(k)}}{4} \right]\sin\left(\left(2\cdot0 + 1\right)x\right)\\
&\qquad - \left[\frac{2a_{k}^{(k)} - a_{k - 1}^{(k)}}{4}\right]\sin\left(\left(2k + 1\right)x\right) 
 - \frac{a_{k}^{(k)}}{4}\sin\left(\left(2\left(k + 1\right) + 1\right)x\right) \\
&= \sum_{i = 0}^{k + 1} a_{i + 1}^{(k + 1)} \sin\left(\left(2i + 1\right)x\right)
\end{align}
Where
$$a_{i + 1}^{(k + 1)} = \begin{cases} 
      \frac{3a_{0}^{(k)} - a_{1}^{(k)}}{4} & i = 0\\
      \frac{2a_{i}^{(k)} - a_{i - 1}^{(k)} - a_{i+1}^{(k)}}{4} & 1\leq i\leq k - 1 \\
      \frac{2a_{k}^{(k)} - a_{k - 1}^{(k)}}{4} & i =  k \\
    -\frac{a_{k}^{(k)}}{4} & i = k + 1
\end{cases}$$
As required. 
