# Showing $\cos (\arcsin(\cos(\theta))) = \lvert \sin (\theta) \rvert$

Where does an absolute function should appear and why?
Having the following equation:

$$\cos (\arcsin(\cos(\theta))) = \lvert \sin (\theta) \rvert$$ With $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.

I draw a triangle and easily found the following:

1. Defining: $\cos(\theta) \equiv \frac{a}{\sqrt{a^2 + 1}} \\$
2. then $\arcsin(\cos(\theta)) = \frac{\pi}{2} - \theta$
3. thus $\cos (\arcsin(\cos(\theta))) = \sin(\theta)$

But why should the absolute value appear and on which phase?
How does the absolute value ultimately surrounds the entire expression?

If we ignored domain and range of functions, we'd end up something like this:

$$\begin{split} \cos(\theta) &= \sin( \frac{\pi}2 + \theta) \\ \arcsin(\cos(\theta)) &= \frac{\pi}2 + \theta (*)\\ \cos(\arcsin(\cos(\theta))) &= \cos(\frac{\pi}2 + \theta) \\ &= -\sin \theta \end{split}$$

However the $(*)$ line isn't quite right. The range of $\arcsin$ is $[-\frac{\pi}2,\frac{\pi}2]$, and the range of the right hand side is $[0,\pi]$. So for positive $\theta$, it's incorrect, it should really be:

$$\begin{split} \arcsin(\cos(\theta)) &= \left\{ \begin{array}{ll} \frac{\pi}2 + \theta & \mbox{if } \theta < 0 \\ \frac{\pi}2 - \theta & \mbox{if } \theta \geq 0 \end{array} \right. \\ &= \frac{\pi}2 - |\theta| \\ \cos(\arcsin(\cos(\theta))) &= \cos(\frac{\pi}2 - |\theta|) \\ &= \left\{ \begin{array}{ll} -\sin(\theta) & \mbox{if } \theta < 0 \\ \sin(\theta) & \mbox{if } \theta \geq 0 \end{array} \right. \\ \end{split}$$

Since $\sin \theta < 0$ for $-\frac{\pi}2 \leq \theta \leq 0$, that simply reduces to $\cos(\arcsin(\cos(\theta))) = |\sin(\theta)|$.

• Thank you, this is very clear to me now. – Dor Jun 4 '15 at 16:15

the left hand side of your equation is an even function of $\theta,$ therefore the right hand side must also be an even function of $\theta.$ that is the reason you see the absolute value sign on the right hand side.

p.s. if you take $0 < t < \pi/2,$ then $\sin^{-1}(\cos t) = \pi/2 - t$ and $\cos(\sin^{-1}(\cos t) ) = \cos(\pi/2 -t) = \sin t = |\sin t|.$

if you take $-\pi/2< t < 0,$ then $\sin^{-1}(\cos t) = \pi/2 + t$ and $\cos(\sin^{-1}(\cos t) ) = \cos(\pi/2 +t) = -\sin t = |\sin t|.$

• Does the absolute function appear as a result of $\arcsin(cos(\theta))$? Whatever happen on those phases - is obscure.. – Dor Jun 4 '15 at 15:57
• @Dor, i have edited my post. see if that helps. – abel Jun 4 '15 at 16:07

Because $\cos$ is positive between $-\frac{\pi}{2}\le \theta \le \frac{\pi}{2}$. Note further that sine is negative between $-\frac{\pi}{2}\le \theta \lt 0$.

• Could you please elaborate? – Dor Jun 4 '15 at 15:45
• @Dor, on left side we have even function, and odd is on left – Michael Galuza Jun 4 '15 at 15:47

$1$. Defining: $\cos(\theta)\equiv\dfrac{a}{\sqrt{a^2+1}}$

This is your first mistake, since we can also have $\cos(\theta)\equiv-~\dfrac{a}{\sqrt{a^2+1}}$

$2$. Then $\arcsin(\cos(\theta))=\dfrac\pi2-\theta$

This is your second mistake, since we can also have $\arcsin(\cos(\theta))=\dfrac\pi2+\theta$

$3$. Thus $\cos(\arcsin(\cos(\theta)))=\sin(\theta)$

This is your third mistake, which follows from the previous two.