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When I go $\sin^{-1}(1)$ I get $90$ degrees but when I put $\frac{1}{\sin(1)}$ in the calculator I get $57.2$ degrees. Why is this?

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    $\begingroup$ $\sin^{-1}x$ is the inverse trigonometric function which gives you the angle $\theta$ for which $\sin\theta=x$ $\endgroup$ – Prasun Biswas Mar 26 '15 at 23:14
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    $\begingroup$ $$\sin^{-1}x\neq \dfrac{1}{\sin x}$$ $\endgroup$ – Prasun Biswas Mar 26 '15 at 23:15
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    $\begingroup$ Actually, a better notation is $\arcsin(x)$ instead of $\sin^{-1}(x)$. $\endgroup$ – Prasun Biswas Mar 26 '15 at 23:16
  • $\begingroup$ So, should I put this as an answer, or are the comments enough? $\endgroup$ – Prasun Biswas Mar 26 '15 at 23:17
  • $\begingroup$ Please see this tutorial on how to format mathematics on this site. $\endgroup$ – N. F. Taussig Mar 27 '15 at 12:12
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$\sin^{-1}(x)$ is confusingly defined, not as $\frac 1{\sin(x)}$ but as the inverse function, the angle $\theta$ such that $\sin (\theta)=x$ (with restrictions to make it a proper function). So $\sin^{-1}(1)$ is the angle with sine of 1, which is $90^\circ$. But if you compute $\sin(1^\circ)$ you get about $0.017452$ and the inverse of this is about $57.2$. Note this is not degrees, it is a pure number. It is numerically close to $\frac {180}\pi$ because for $x \ll 1$ in radians, $\sin x \approx x$, so $\sin 1^\circ=\sin (\frac \pi{180})$radians $\approx \frac \pi{180}$

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