$\arcsin(\sin(5\pi/4))$
Having trouble with this question
$\arcsin(\sin(5\pi/4))$
wouldn't the arcsin and sin cancel each other out, therefor this would just equal $5\pi/4$. the answer in the back of my book is $-\pi/4$.
$\endgroup$ 24 Answers
$\begingroup$We have $\sin(\arcsin x)=x$ for all $x\in[-1,1]$, and $\arcsin(\sin x)=x$ for all $x\in[-\pi/2,\pi/2]$. The key, important difference is that $\sin(\arcsin x)$ is only defined for $x\in[-1,1]$, while $\arcsin(\sin x)$ is defined for all real $x$. The latter's unlimited domain is the source of many a mistake.
$\endgroup$ $\begingroup$$\arcsin(x)$ is a number whose sine is $x$. But there are a lot of numbers whose sine is $x$, so which one is $\arcsin(x)$? The values of $\sin(x)$ are generally between $-\pi/2$ and $\pi/2$, which ensures that for every $x$ (between $-1$ and $1$), there is exactly one answer. So the question is asking which angle between $-\pi/2$ and $\pi/2$ has a sine equal to $\sin(5\pi/4)$?
$\endgroup$ $\begingroup$$\sin\left( \frac{5\pi}{4} \right)= - \sqrt{\frac12}$
$\arcsin \left( - \sqrt{\frac12} \right) = -\frac{\pi}{4}$ if you are using the usual definition of $\arcsin: [-1,1] \to\left[-\frac{\pi}2,\frac{\pi}2\right]$. As a function, $\arcsin$ can only give a single value
Graphically:
(The range of arcsin is usually taken to be $[\frac{-\pi}{2}, \frac{\pi}{2}]$.)
$\textrm{sin}(5 \pi /4)=\textrm{sin}(-\pi/4)$, so you have $\textrm{sin}^{-1}(\textrm{sin}(5\pi /4))=\textrm{sin}^{-1}(\textrm{sin}(-\pi/4))=-\pi/4$
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