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Given $A,B,a,b$ are constants and x are variable. $A,B$ don't equal $0$.

Does $A\sin(x + a) = B\sin(x + b)$ for all $x$ implies $a = b + 2n\pi$?

I only managed to show that $a-b = n\pi$ by putting $x = -b$.

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    $\begingroup$ Only if $A$ and $B$ have the same sign. $\endgroup$ – Kitegi May 13 '15 at 13:18
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You have:

  • taking $x = 0: A\sin a = B\sin b$
  • taking $x = \frac \pi 2: A\cos a = B\cos b$

Hence $A^2 = B^2$, or $A = \pm B$. Putting in the first equation: $$ \sin a = \sin b \implies \exists k: (a = b + 2k\pi \text{ or } a = \pi - b + 2k\pi ) $$ or $$ \sin a = -\sin b \implies \exists k: (a = -b + 2k\pi \text{ or } a = \pi + b + 2k\pi ) $$

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