This page demonstrates how to use differentiation to prove the identity: \[ \arcsin(x) + \arccos(x) = \dfrac{\pi}{2} \] for all values of \( x \) in the domain \([-1, 1]\).
Define the function: \[ f(x) = \arcsin(x) + \arccos(x) \] We will show that \( f(x) \) is constant by computing its derivative.
The derivative of \( f(x) \) is: \[ f'(x) = \dfrac{d}{dx}[\arcsin(x)] + \dfrac{d}{dx}[\arccos(x)] = \dfrac{1}{\sqrt{1 - x^2}} - \dfrac{1}{\sqrt{1 - x^2}} = 0 \] Since \( f'(x) = 0 \) for all \( x \in (-1, 1) \), the function \( f(x) \) is constant.
To determine the value of this constant, we evaluate \( f(x) \) at a specific point. Let’s use \( x = 0 \): \[ f(0) = \arcsin(0) + \arccos(0) = 0 + \dfrac{\pi}{2} = \dfrac{\pi}{2} \] Therefore, \[ \arcsin(x) + \arccos(x) = \dfrac{\pi}{2} \] for all \( x \in [-1, 1] \).
For more examples and applications of derivatives, visit our page on applications of differentiation.