Any hints on how to simplify this $\arcsin(x) + \arcsin(\cos(\arcsin(x)))$ to a single term?

Question

$\arcsin(x) + \arcsin(\cos(\arcsin(x)))$

Completely stupefied about where to start, if I'm being honest.

Answer 1

Try drawing a triangle.

Alternatively, you can start by use the identity $\cos^2(x)=1-\sin^2(x)$. So $\cos(\arcsin(x))=\sqrt{\cos^2(\arcsin(x))}=\sqrt{1-\sin^2(\arcsin(x))}=\sqrt{1-x^2}$

Answer 2

Let $\theta=\arcsin x$.

If we assume that $0\le \theta\le\frac\pi 2$, then $\cos\theta=\sin(\frac\pi 2-\theta)$ and: \begin{aligned}\arcsin x + \arcsin(\cos(\arcsin x)) &= \theta + \arcsin(\cos \theta)\\ &= \theta + \arcsin\Big(\sin\big(\frac\pi 2 - \theta\big)\Big)\\ &= \theta + \big(\frac\pi 2 - \theta\big)\\ &=\frac \pi 2 \end{aligned}

Answer 3

If $-1 \le x \le 1$ then $\arcsin x$ is going to be interpreted as being an angle $\arcsin x=\theta$ between $-\frac \pi2\le \theta \le \frac \pi 2$.

If $x\ge 0$ then $\theta \ge 0$ and $\sin \theta$ and $\cos \theta$ are both $\ge 0$ and $\cos \theta = \sin (\frac \pi 2- \theta)$

If $x < 0$ then $-\frac \pi 2\le \theta \le 0$ and $\sin \theta \le 0$ and $\cos \theta \ge 0$ and $\cos \theta =\sin (\frac \pi 2 + \theta)$.

In either case $\cos \theta \ge 0$ and $\arcsin (\cos \theta) = \begin{cases}\frac \pi 2-\theta& \theta \ge 0\\\frac \pi 2 + \theta& \theta \le 0\end{cases}$

And $\arcsin(x) + \arcsin(\cos(\arcsin(x)))=\begin{cases}\theta +\frac \pi 2-\theta=\frac \pi 2& \theta \ge 0\\\theta+\frac \pi 2 + \theta=2\theta +\frac \pi 2=2x+\frac \pi 2& \theta \le 0\end{cases}$

Let's double check.

Let $x =\frac 12$. $\sin \frac \pi 6 = \frac 12$ so $\arcsin \frac 12 = \frac \pi 6$. $\cos \frac \pi 6=\frac {\sqrt 3}2$ and $\sin \frac \pi 3 = \frac {\sqrt{3}}2$ so $\arcsin(\cos(\arcsin x))= \frac \pi 3$. And $\arcsin x + \arcsin(\cos(\arcsin x)) = \frac \pi 6 + \frac \pi 3 =\frac \pi 2$. That confirms.

Le $x = -\frac 12$. $\sin-\frac \pi 6=-\frac 12$ so $\arcsin -\frac 12=-\frac \pi 6$. $\cos(-\frac \pi 6) = \frac {\sqrt 3}2$ and so, like above, $\arcsin(\cos(\arcsin x)) =\frac \pi 3$. And so $\arcsin x + \arcsin(\cos(\arcsin(x))= -\frac \pi 6 +\frac \pi 3 = \frac \pi 6$. Which $= 2(-\frac \pi 6) + \frac \pi 2$. That confirms.

Answer 4

We have $\cos(\arcsin x)=\sqrt{1-x^2}$ which can be seen by drawing a right triangle with angle $\theta=\arcsin(x)\Rightarrow \sin\theta=x$. We also have the identity (see here) $$\arcsin(\alpha)+\arcsin(\beta)=\arcsin\left(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}\right)$$ which holds for $\alpha^2+\beta^2\leq 1$. Putting the two together, the expression simplifies to $$\arcsin\left(x|x|+1-x^2\right) $$ We can note that when $x\in[0,1]$, this is $\arcsin(1)=\pi/2$. When $x\in[-1,0)$, it becomes $\arcsin(1-2x^2)$.