# Showing that $\arcsin x + \arccos y = \frac{\pi}{2}$ if and only if $x = y$

I was just wondering about this identity:

$$\arcsin x + \arccos x = \frac{\pi}{2} .$$

That a thought came to my mind that in general $$\arcsin x + \arccos y = \frac{\pi}{2} \qquad \textrm{if and only if} \qquad x = y .$$ I have a hunch that it's true, and I have kind of done a self-satisfactory but illegal proof by using hit and trial, and later I also tried using graphs, but I am still stuck.

## 4 Answers

$$arcsin (x)+arccos(y)=\frac{\pi}{2}$$ $$\Longleftrightarrow arccos(y)=\frac{\pi}{2}-arcsin (x)$$ $$\Longleftrightarrow arccos(y)=arccos(x)$$ $$\Longleftrightarrow y=x$$ The last step follows from the fact that $$arccos(x)$$ is invertible.

Hope it helps:)

• yeah it helps a lot...... i feel so dumb cuz i was just involved in graphical analysis and like kinda did not really go into algebraic analysis..... Thanks alot nice meetin you – arnav009 Apr 6 at 5:21

Hint: Yes, if $$\sin^{-1}{x} + \cos^{-1}{y} = \pi/2$$ (where $$x,y\in [-1,1]$$), then $$x=y$$. To show this, suppose that $$\sin^{-1}{x} + \cos^{-1}{y} = \pi/2$$ (where $$x,y\in [-1,1]$$). Then $$\sin^{-1}{x} = \pi/2 - \cos^{-1}{y}$$. Now take the sine of both sides and recall some trigonometric identities. In particular, what are $$\sin\left(\sin^{-1}{x}\right)$$, $$\sin(\pi/2-\theta)$$, and $$\cos\left(\cos^{-1} y\right)$$?

Conversely, if $$x=y$$ (where $$x,y\in [-1,1]$$), then $$\sin^{-1}{x} + \cos^{-1}{y} = \pi/2$$. This is a standard result. See e.g. here for a proof.

Hint For the remaining direction, use the identity $$\arcsin x + \arccos x = \frac{\pi}{2}$$ you mention together with the fact that $$x \mapsto \arcsin x$$ is strictly increasing.

If $$x=y$$, then consider a right triangle where one leg is $$1$$ unit long and the other leg is $$x$$ units long. Then $$\arcsin(x)+\arccos(x)$$ is just adding the two non-right angles, so that sum will be $$\frac{\pi}{2}$$. And since we assumed $$x=y$$ here, then the conclusion is $$\arcsin(x)+\arccos(y)=\frac{\pi}{2}$$.

Now assume $$\arcsin(x)+\arccos(y)=\frac{\pi}{2}$$. Apply $$\cos$$ to both sides: \begin{align} \cos\mathopen{}\left(\arcsin(x)+\arccos(y)\right)\mathclose{}&=\cos\pi/2\\ \cos(\arcsin(x))\cos(\arccos(y))-\sin(\arcsin(x))\sin(\arccos(y))&=0\\ \sqrt{1-x^2} y - x\sqrt{1-y^2}&=0\\ \sqrt{1-x^2} y &= x\sqrt{1-y^2}\\ \end{align} At this point, note that either the sign of $$y$$ and $$x$$ must be the same (which happens when the square roots are positive) or $$\lvert x\rvert=\lvert y\rvert = 1$$ (which happens when the square roots are $$0$$). Now square both sides. \begin{align} \left(1-x^2\right)y^2 &= \left(1-y^2\right)x^2\\ y^2-x^2y^2&=x^2-x^2y^2\\ y^2&=x^2 \end{align} So given our note in the middle of that equation, either $$y=x$$ or ($$x=1$$ and $$y=-1$$) or ($$x=-1$$ and $$y=1$$). But ($$x=1$$ and $$y=-1$$) makes $$\arcsin(x)+\arccos(y)=3\pi/2$$. And ($$x=-1$$ and $$y=1$$) makes $$\arcsin(x)+\arccos(y)=-\pi/2$$. So it must be that $$y=x$$.

• thanks alot ...... i really like your style and contradictions... – arnav009 Apr 6 at 5:23