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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.

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

Hope it helps:)

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  • $\begingroup$ 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 $\endgroup$ – arnav009 Apr 6 at 5:21
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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.

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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.

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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 $y=x$.

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

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