How can I simplify $\sum_{i=1}^{n} \sin(x - y_i)$ to the form $Asin(x-y_0)$?

Question

For example for $n=2$ the new phase $y_0$ is the average of $y_1$ and $y_2$ but this dowsn't seem to be true in general.

And what about simplifying $\sum_{i=1}^{n} A_i sin(x - y_i)$ ?

Analytically it's easy to see why we get one sine. But is there an elementary reasoning for this?

Answer 1

$$ \sum_j^n A_j\sin(x-y_j)={\bf Im}\sum_j^n A_je^{ix-iy_j}={\bf Im}\left(\sum_j^n A_je^{-iy_j}\right)e^{ix}={\bf Im}\,re^{ix-i\vartheta}=r\sin(x-\vartheta), $$ where $$ \sum_j^n A_je^{-iy_j}=re^{i\vartheta} $$

Answer 2

Let $\sum\limits_{i=1}^n\cos y_i=a$ and $\sum\limits_{i=1}^n\sin y_i=b$.

Thus, $$\sum_{i=1}^n\sin(x- y_i)=a\sin x-b\cos x=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin x-\frac{b}{\sqrt{a^2+b^2}}\cos x\right).$$ Can you end it now?