Hide answer & extension
Let \(y = \arccos(x)\)
This means that \(x = \cos(y)\)
This gives us this triangle, as cosine is adjacent ÷ hypotenuse.
Call the angle at the top of the triangle \(z\).
Sine is opposite ÷ hypotenuse, so \(\sin(z) = x\)
So \(z = \arcsin(x)\)
Angles in triangle add to \(\pi\), so \(z = \frac{\pi}{2}-y\)
Hence, \(\arcsin(x)= \frac{\pi}{2}-y\)
Therefore \(\arccos(x) + \arcsin(x) = y + \frac{\pi}{2}-y\) \(=\frac{\pi}{2}\)
Extension
What are the values of \(\mathrm{arcsec}(x) + \mathrm{arccosec}(x)\) and \(\arctan(x) + \mathrm{arccot}(x)\)?
![](javascript:showlimage("arccos-triangle.png"))
