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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?

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    Do you understand what even and odd functions are? If you can't recall, search Wikipedia, or google it. Do you understand that trigonometric values of an angle may differ depending on the quadrant(s) in which its terminal side falls? Look it up. Think.... – 2017-01-15
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    Yes , I've already seen all lectures related to trigonometry , except hyperbolic trigonometry and the inverse trigonometric functions , – 2017-01-15
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    None of the 17 questions you have asked has an accepted answer. With all due respect, do you think any one of them helpful to you? – 2017-01-15
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    you may not believe me , but until know I didn't know that there was an "accepeted answer button" ,my sincere appologies to everyone that has gave me an answer and didn't receive a reward , I'am going to repair that . – 2017-01-15

2 Answers 2

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$\sin(\arcsin(-x))=-x, \sin(-\arcsin(x))=-\sin(\arcsin(x))=-x$, 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 $\arcsin(2x)=\pi-2x$.

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    ok and for the second one ? – 2017-01-15
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    Hilbert, please be polite, and not demanding in your tone. – 2017-01-15
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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.