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Let $a$ be a given number. Determine all numbers $x$ such that $\sin x = \sin a$.

You may suppose that $0 \le a \lt 2\pi$, and distinguish the cases $a = \frac\pi2$, $a = \frac{-\pi}2$ and $a \neq \frac{\pm\pi}2$


I know that I have to find a formula that gives me all numbers $x$ such that sin $x = \sin a$. Like $\sin \pi/6 = \sin 5\pi/6 = \sin 13\pi/6$ and so on. But I am having problem to find that.

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  • $\begingroup$ Have you put any effort into finding a solution at all? $\endgroup$ – TomGrubb May 22 '15 at 21:06
  • $\begingroup$ Yes. I know that I have to find a formula that gives me all numbers x such that sin x = sin a. Like sin pi/6 = sin 5pi/6 = sin 13pi/6 and so on. But I am having problem to find that. $\endgroup$ – LucasPG May 22 '15 at 21:11
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Hint: $$\sin(\theta+2\pi k)=\sin \theta,k\in\Bbb{Z}$$ Also $$\sin(\theta)=\sin(\pi-\theta)$$

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  • $\begingroup$ Ok. So I have $\sin a = \sin (a+2\pi * k)$ and $\sin a = \sin (\pi − a + 2\pi * k)$. But I am having problems to join them, so I would have just one function that gives me all numbers x that satisfy the equation $\sin x = \sin a$. Can you help me? $\endgroup$ – LucasPG May 22 '15 at 21:31
  • $\begingroup$ @LucasPG Why would you need 1 full equation? $\endgroup$ – Cyclohexanol. May 22 '15 at 21:36
  • $\begingroup$ Well, the exercise was asking to determine all numbers x. So I thought there would be a single equation that would give me all the numbers x. But if it is not possible to do that, then I think that is the answer the exercise wants. Do I need to do more than this, or is it just enough? $\endgroup$ – LucasPG May 22 '15 at 21:43
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Consider the diagram below:symmetries_of_sine_and_cosine

Since reflection in the $y$-axis does not change the value of the sine function, $\sin(\pi - \theta) = \sin\theta$. Since coterminal angles have the same sine, $\sin(\theta + 2k\pi) = \sin\theta$ for each $k \in \mathbb{Z}$. Putting the two observations together yields $\sin(\pi - \theta + 2k\pi) = \sin\theta$ for each $k \in \mathbb{Z}$.

If $\sin x = \sin a$, then \begin{align*} x & = \begin{cases} a + 2k\pi, k \in \mathbb{Z}\\ \pi - a + 2k\pi, k \in \mathbb{Z}\\ \end{cases}\\ & = \begin{cases} a + 2k\pi, k \in \mathbb{Z}\\ -a + (2k + 1)\pi, k \in \mathbb{Z}\\ \end{cases}\\ & = (-1)^na + n\pi, n \in \mathbb{Z} \end{align*}

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