Why Is \cos y Positive When Differentiating \arcsin x?

[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}?

 

[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.