# Differentiation of inverse trig

1. Feb 26, 2014

### perishingtardi

Say we want to differentiate $\arcsin x$. To do this we put $$y=\arcsin x.$$ Then $$x=\sin y \implies \frac{dx}{dy}= \cos y.$$ Then we use the relation $$\sin^2 y + \cos^2 y = 1 \implies \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}.$$ Therefore $$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}.$$

My question is that when we use $$\sin^2 y + \cos^2 y = 1,$$ how do we know that $$\cos y = \sqrt{1 - \sin^2 y}$$ rather than $$\cos y = -\sqrt{1 - \sin^2 y}$$?

2. Feb 26, 2014

### eigenperson

Because arcsine, by convention, always returns angles in the range $[-\pi/2, \pi/2]$. This guarantees that $y = \arcsin x$ is in that range, so $\cos y$ has to be positive. Therefore, you know it must be the positive square root, instead of the negative one.