Prove that $\cos^{-1}(-x) = \pi - \cos^{-1}x, \; x \in [-1, 1]$.

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> [!TLDR]
> With θ = cos⁻¹x ∈ [0, π], cos(π−θ) = −x and π−θ ∈ [0,π], so cos⁻¹(−x) = π − θ.
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> We prove cos⁻¹(−x) = π − cos⁻¹x for x ∈ [−1, 1].
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> Step 1: Let cos⁻¹x = θ. Then cosθ = x and θ ∈ [0, π] (principal range).
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> Step 2: Consider cos(π − θ) = −cosθ = −x.
> …

Inverse function	Domain (allowed inputs $x$ )	Principal value branch (angles returned)
$\sin^{-1} x$	$[-1,1]$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\cos^{-1} x$	$[-1,1]$	$[0,\pi]$
$\tan^{-1} x$	$\mathbb{R}$	$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\cot^{-1} x$	$\mathbb{R}$	$(0,\pi)$
$\sec^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$[0,\pi]\setminus\{\tfrac{\pi}{2}\}$
$\csc^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$

List - I	List - II
A) Domain of $\sin^{-1} x$	i) $\left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$
B) Range of $\tan^{-1} x$	ii) $[0, \pi]$
C) Range of $\cos^{-1} x$	iii) $[-1, 1]$

Q.Prove that $\cos^{-1}(-x) = \pi - \cos^{-1}x, \; x \in [-1, 1]$ .

Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

The idea: restrict to one clean interval

Why the intervals differ …

Concept and Intuition

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Q.Prove that cos⁡−1(−x)=π−cos⁡−1x, x∈[−1,1]\cos^{-1}(-x) = \pi - \cos^{-1}x, \; x \in [-1, 1]cos−1(−x)=π−cos−1x,x∈[−1,1].

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Q.Prove that $\cos^{-1}(-x) = \pi - \cos^{-1}x, \; x \in [-1, 1]$ .

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