Prove the following identities:

\begin{align} (a) && \sin(\arccos(y)) &= \sqrt{1-y^2}\\ (b) && \arcsin(\cos(x)) &= \frac{\pi}{2}-x \end{align}

For (a) I am not sure how I would get a root out of any identity. For (b) I can transform the $\cos$ to $\sin(x + \pi / 2)$ so I would assume the result will be $x + \pi / 2$, and not $\pi / 2 - x$.

  • $\begingroup$ Hint: $\sin A=\sqrt{1-\cos^2A}$. What if you substitute something into that? $\endgroup$ – Akiva Weinberger Dec 16 '15 at 20:07
  • $\begingroup$ @AkivaWeinberger It's only true if $A\in[2\pi k,\pi+2\pi k]$ for some $k\in\mathbb Z$. $\endgroup$ – user236182 Dec 16 '15 at 20:08
  • $\begingroup$ @user236182 True. But $A$ is in that range (when you figure out what to set $A$ equal to). $\endgroup$ – Akiva Weinberger Dec 16 '15 at 20:11
  • $\begingroup$ $\cos x=\sin\left(\frac{\pi}{2}-x\right)$, but $\sin\left(\frac{\pi}{2}+x\right)=-\cos x$. $\endgroup$ – user236182 Dec 16 '15 at 20:12
  • $\begingroup$ Identity (a) holds for all $y$ for which the two sides make sense (as real numbers), namely $-1\le y\le1$, but identity (b) does not hold for all $x$. $\endgroup$ – Barry Cipra Dec 16 '15 at 20:15

For 'a':

$$\sin\left(\arccos(y)\right)=\sqrt{1-\cos^2\left(\arccos(y)\right)}=$$ $$\sqrt{1-\cos\left(\arccos(y)\right)\cos\left(\arccos(y)\right)}=\sqrt{1-yy}=\sqrt{1-y^2}$$

HINT (for 'b'):

$$\arcsin(\cos(x))\ne\frac{\pi}{2}-x$$ enter image description here



$\arccos(y)=x$ means that $\cos x= y$

so $\sin (\arccos (y))=\sin x$ and $\sin x=\sqrt{1-\cos^2 x}$

You can do the same for the other identity:

$\arcsin(\cos x)=y \iff \sin y=\cos x$, than use $\sin(\pi/2 -x)=\cos x$.


This answer pertains to part (b).

If it were true that $\arcsin(\sin\theta)=\theta$ for all $\theta$, then the fact that $\cos x=\sin(x+\pi/2)$ would imply $\arcsin(\cos x)=\arcsin(\sin(x+\pi/2))=x+\pi/2$, as the OP observed. But $\arcsin(\sin\theta)$ does not equal $\theta$ for all $\theta$. Instead we have

$$\arcsin(\sin\theta)=\theta\quad\text{if and only if }-\pi/2\le\theta\le\pi/2$$

What's actually going on here is that identity (b) has been stated incompletely. It should say

$$\arcsin(\cos x)=\pi/2-x\quad\text{for }0\le x\le\pi$$

Note that if we let $\theta=\pi/2-x$, then

$$0\le x\le\pi\implies\pi/2\le\theta\le\pi/2$$ and

$$\sin\theta=\sin(\pi/2-x)=\cos x$$

Putting these together, we have

$$\arcsin(\cos x)=\arcsin(\sin\theta)=\theta=\pi/2-x$$

but only for $0\le x\le\pi$.


The value of $\arccos y$ is in the interval $[0,\pi]$, where the sine is positive. So, if $\alpha=\arccos y$, we know that $\cos\alpha=y$ and $$ \sin\arccos y=\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-y^2} $$ Similarly, $\arcsin$ has values in the interval $[-\pi/2,\pi/2]$. If $\alpha=\arcsin\cos x$, then $-\pi/2\le\pi/2$ and $$ \sin\alpha=\cos x $$ so $$ \cos\left(\frac{\pi}{2}-\alpha\right)=\cos x $$ Note that $$ 0\le\frac{\pi}{2}-\alpha\le\pi $$ Thus, if $x\in[0,\pi]$, we can conclude that $$ \arcsin\cos x=\frac{\pi}{2}-x $$ Otherwise the relation is not true. For instance, if $x=\frac{3\pi}{2}$, we have $\cos x=0$ and so $$ \arcsin\cos\frac{3\pi}{2}=\arcsin0=0\ne\frac{\pi}{2}-\frac{3\pi}{2}=-\pi $$


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