<s>[INST] Adding sine waves of different phase, sin(π/2) + sin(3π/2)? Adding sine waves of different phase, what is $\sin(\pi/2) + \sin(3\pi/2)$?
Please could someone explain this.
Thanks.
 [/INST] $\sin(\pi+x)=\sin \pi\cos x+\cos\pi\sin x=-\sin x$ as $\sin \pi=0,\cos\pi=-1$
This can be achieved directly using "All-Sin-Tan-Cos" formula or using $\sin 2A+\sin 2B=2\sin(A+B)\cos(A-B)$, $\sin(\pi+x)+\sin x=2\sin\left(x+\frac \pi 2\right)cos\left(\frac \pi 2\right)=0$
So, $\sin(x)+\sin(\pi+x)=0$
More generally, $\sin(x+c)+\sin(x+d)=2\sin\left(x+\frac{c+d}2\right)\cos\left(\frac{c-d}2\right)$
So, the resultant phase is the arithmetic mean of the phases of the original waves provided $\cos\left(\frac{c-d}2\right)\ne 0,$ i.e., $\frac{c-d}2\ne \frac{(2n+1)\pi}2$ or $c-d\ne(2n+1)\pi$ where $n$  is any integer.
Here, $c-d=0-\pi=-\pi\implies \cos\left(\frac{c-d}2\right)=0 $
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