Why does $\arcsin(-x) = -\arcsin(x)$? [closed]

$\begingroup$

What is the property that tells us $\arcsin(-x) = -\arcsin(x)$ ?

I've also seen in an exercise that : $\arcsin (\sin(2x)) = 2x ~~$ if $x \in\left[0,\frac{\pi}{4}\right]$

And what about: $\arcsin(\sin(2x)) = \pi - 2x ~~$ if $x \in \left]\frac{\pi}{4},\frac{\pi}{2}\right]$?

What justifies these relations?

$\endgroup$ 3

2 Answers

$\begingroup$

$\sin(\arcsin(-x))=-x, \sin(-\arcsin(x))=-\sin(\arcsin(x))=-x$, since $\sin:[-\pi/2,\pi/2]\rightarrow [-1,1]$ is the inverse of $\arcsin$ we deduce that $\arcsin(-x)=-\arcsin(x)$ since the restriction of $\sin$ on $[-\pi/2,\pi/2]$ is injective.

If $x\in [\pi/4,\pi/2], 2x\in [\pi/2,\pi]$ and $\pi-2x\in [0,\pi/2]$, since $\sin(\pi-2x)=\sin(2x)$ and $\sin$ is the inverse of $\arcsin:[-1,1]\rightarrow [-\pi/2,\pi/2]$, we deduce that $\pi-2x$ is the unique element of $[-\pi/2,\pi/2]$ such that $\sin(\pi-2x)=\sin(2x)$ and $\arcsin(2x)=\pi-2x$.

$\endgroup$ 1 $\begingroup$

We can also write: $$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$ So, if you're more familiar with the exponential function, it can b easier to derive identities with this.

$\endgroup$