I was reading a book about spectral analysis in time series and found the following statement without proof.
Statement
The function $f(r)=a_2^{r/2}\frac{(\sin((r+1)\theta)-a_2\sin((r-1)\theta))}{(1+a_2)\sin(\theta)}$
Can be written as
$$f(r)=a_2^{r/2}\frac{\sin(r\theta+\psi)}{\sin (\psi)}$$
where $\tan(\psi)=\tan(\theta)\frac{1+a_2}{1-a_2}$
Anyone has an idea about how to prove this result?
$$\begin{align}&\frac{\sin((r+1)\theta)-a_2\sin((r-1)\theta)}{(1+a_2)\sin\theta}=\\&=\frac{(\sin(r\theta)\cos\theta+\cos(r\theta)\sin\theta)-a_2(\sin(r\theta)\cos\theta-\cos(r\theta)\sin\theta)}{(1+a_2)\sin\theta}\\ &=\frac{(1-a_2)\sin(r\theta)\cos\theta+(1+a_2)\cos(r\theta)\sin\theta}{(1+a_2)\sin\theta}\\ &=\sin(r\theta)\cot\theta\frac{1-a_2}{1+a_2}+\cos(r\theta)\\ &=\sin(r\theta)\cot\psi+\cos(r\theta)\\ &=\frac{\sin(r\theta)\cos\psi+\cos(r\theta)\sin\psi}{\sin\psi}\\ &=\frac{\sin(r\theta+\psi)}{\sin\psi} \end{align}$$