This page demonstrates how to use differentiation to prove the important trigonometric identity: \[ \arcsin(x) + \arccos(x) = \dfrac{\pi}{2} \] for all values of \( x \) in the domain \([-1, 1]\).
Overview of the Method
To prove this identity, we define a helper function: \[ f(x) = \arcsin(x) + \arccos(x) \] We will show that \( f(x) \) is constant across its domain by computing its first derivative and demonstrating that it equals zero everywhere on the open interval \((-1, 1)\).
Step-by-Step Proof
Step 1: Compute the First Derivative
Recall the standard derivative formulas for inverse sine and inverse cosine:
- \( \dfrac{d}{dx}[\arcsin(x)] = \dfrac{1}{\sqrt{1 - x^2}} \)
- \( \dfrac{d}{dx}[\arccos(x)] = -\dfrac{1}{\sqrt{1 - x^2}} \)
Differentiating \( f(x) \) term by term gives: \[ f'(x) = \dfrac{d}{dx}[\arcsin(x)] + \dfrac{d}{dx}[\arccos(x)] = \dfrac{1}{\sqrt{1 - x^2}} - \dfrac{1}{\sqrt{1 - x^2}} = 0 \]
Step 2: Establish that the Function is Constant
Since \( f'(x) = 0 \) for all \( x \in (-1, 1) \), it follows from the Mean Value Theorem (or properties of constant functions) that \( f(x) \) must be a constant value, say \( C \), throughout the interval. By continuity, this constancy extends to the closed endpoints \([-1, 1]\).
Step 3: Evaluate at a Specific Point
To determine the exact value of this constant \( C \), we evaluate \( f(x) \) at an easily computable point within the domain. Let’s choose \( x = 0 \): \[ f(0) = \arcsin(0) + \arccos(0) \] Since \( \arcsin(0) = 0 \) and \( \arccos(0) = \dfrac{\pi}{2} \): \[ f(0) = 0 + \dfrac{\pi}{2} = \dfrac{\pi}{2} \]
Final Result
Because \( f(x) \) is a constant function equal to \( \dfrac{\pi}{2} \) everywhere on its domain, we conclude that: \[ \arcsin(x) + \arccos(x) = \dfrac{\pi}{2} \] for all \( x \in [-1, 1] \).