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Adding sine waves of different phase, what is $\sin(\pi/2) + \sin(3\pi/2)$?

Please could someone explain this.

Thanks.

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Whatever else you are doing, $\sin(\pi/2)+\sin(3\pi/2)$ is not the result of adding two sine waves of different phase; it is the result of adding two constants, and the sum happens to be $0$ as lab bhattacharjee shows you in the answer you have accepted. The sum of two sine waves (of the same frequency) but different phases would be $\sin(\omega t + \theta_1) + \sin(\omega t + \theta_2)$ which would also be $0$ if you chose $\theta_1$ and $\theta_2$ to be $\pi/2$ and $3\pi/2$, but is a sinusoid of the same frequency but different amplitude and phase in general. –  Dilip Sarwate Oct 31 '12 at 16:16
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2 Answers 2

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$\sin(\pi+x)=\sin \pi\cos x+\cos\pi\sin x=-\sin x$ 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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Heres the plot for $\sin(L)$ where $L$ goes from $(0, \pi/2)$

enter image description here

Heres the plot for $\sin(L) + \sin(3L)$ where $L$ goes from $(0, \pi/2)$

enter image description here

I hope this distinction is useful to you. This was done in Mathematica.

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