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I'm looking to get a particular answer in the form of $\sin(z)$, and I managed to reach an answer in the form $\sin(x)/\sin(y)$. I've checked on a calculator which has confirmed that they're the same number, but how can I convert the fraction into a single sin in order to show that without relying on the calculator?

Winther
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Dalekcaan1963
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    As the current answer explains, a general method does not exist. Whatever can be done to show the equality you have will require more details about $x$, $y$, and $z$. – Eric Towers Oct 17 '16 at 05:39
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    Both answers below (understandably) assume that $x$ and $y$ are real, and that $z$ has to be real, and in that case a general method does not exist as explained. But the question would be much more interesting if it would focus on complex numbers... –  Oct 17 '16 at 09:51
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    This is definately not what you are looking for (as this is tagged trigonometry) but if we allow $z$ to be a complex number then there do exist solutions. For example taking $z = -i\log\left(i\omega \pm \sqrt{1-\omega^2}\right)$ where $\omega = \frac{\sin(x)}{\sin(y)}$ then $\sin(z) = \frac{\sin(x)}{\sin(y)}$ (apart from the values where $\sin(y) = 0$). – Winther Oct 17 '16 at 16:46

2 Answers2

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There is no general solution $\sin(x)/\sin(y) = \sin(z)$ (for all $x,y$). This can easily be seen by setting $y=0$ and then the fraction diverges - which $\sin(z)$ would never do.

Simply Beautiful Art
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Penguino
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    Whoa, what a sneaky little answer. – Simply Beautiful Art Oct 17 '16 at 00:47
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    Finally, a nice answer I can understand on Mathematics SE. – camden_kid Oct 17 '16 at 08:05
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    The question states that they already have the form $\sin x / \sin y$, so it is implied that we don't need a general solution, only one for when $\sin y \neq 0$. – JiK Oct 17 '16 at 11:54
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    @JiK Even so, it's easy to imagine $\sin(y)$ being very small, so $\sin(x)/\sin(y)$ being very large but defined. The point is that with the information given in the question there is no general method. – Richard Rast Oct 17 '16 at 13:19
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You will always have $-1\leq\sin z\leq 1$, while $\sin x/\sin y $ can be any real number or even be undefined when $y=0$. For example, $$\frac {\sin\pi/2}{\sin\pi/4}=\frac1 {1/\sqrt2}=\sqrt2>1$$ and no choice of $z $ will give you $\sin z=\sqrt2$.

In the case in which the quotient is between $-1$ and $1$, you can write $$z=\arcsin\left (\frac {\sin x}{\sin y}\right), $$ but I don't think there is a simpler expression.

Martin Argerami
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