Two different angles whose amplitudes are in the interval $[0,2\pi]$ have the same sine. What is the sum of their amplitudes?

A. $\pi$ $$\\$$ B. A multiple of $\pi$ $$\\$$ C. A multiple of $2\pi$ $$\\$$ D. $2\pi$

I know that the amplitudes of these angles have to be $\alpha$ and $\pi-\alpha$ and so

$$(\pi-\alpha)+\alpha = \pi$$

But my book says the solution is B. I can't understand how that is though, as I couldn't seem to find any 2 diferent anges of amplitude within the given interval for which the sum would be a multiple of $\pi$.

Can someone explain this to me?

  • $\begingroup$ Have the same sine as in exactly the same or same numerical value disregarding negative signs. Because $\pi/4$ and $3\pi/4$ produce the same answer in sine and are $\pi/2$ apart $\endgroup$ – Triatticus May 2 '17 at 18:56
  • $\begingroup$ The angles could be bigger than $\pi$. $\endgroup$ – Lord Shark the Unknown May 2 '17 at 18:58
  • 1
    $\begingroup$ $0$ and $2\pi$ have the same sine. $\endgroup$ – lulu May 2 '17 at 19:00
  • $\begingroup$ If your angles are $\sin \frac {5\pi}{4} = \sin \frac {7\pi}{4} = -\frac {\sqrt {2}}{2}$ So $3\pi$ is in the solution set. $\endgroup$ – Doug M May 2 '17 at 19:10

I know that the amplitudes of these angles have to be $\alpha$ and $\pi-\alpha$ ...

That is not true. The angles could be $\pi + \alpha$ and $2\pi - \alpha$, where $0\leq \alpha \leq \frac\pi2.$ Then the sine of either $\pi + \alpha$ or $2\pi - \alpha$ will be zero (if $\alpha = 0$) or negative.


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