# Relation f⁻¹(x) with f⁻¹(-x) – Theorems

## Relation f⁻¹(x) with f⁻¹(-x) – Theorems

Theorem: $${{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x\ \ for\ \ all\ \ x\in \left[ -1,1 \right]$$.

Proof: $$-x\in \left[ -1,1 \right]\ \ for\ \ all\ \ x\in \left[ -1,1 \right]$$,

Let sin⁻¹(-x) = θ . . . (1)

-x = sinθ

x = -sinθ

x = sin(-θ)

we know that x ϵ [-1, 1] and  $$-\theta \in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]\ \ for\ \ all\ \ \theta \in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]$$,

-θ =sin⁻¹(x) . . . (2)

From equation (1) and (2)

sin⁻¹(-x) = θ

-θ =sin⁻¹(x)

⇒ θ = -sin⁻¹(x)

sin⁻¹(-x) = -sin⁻¹(x)

Hence proved

Theorem: $${{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}x\ \ for\ \ all\ \ x\in \left[ -1,1 \right]$$.

Proof: $$-x\in \left[ -1,1 \right]\ \ for\ \ all\ \ x\in \left[ -1,1 \right]$$,

Let cos⁻¹(-x) = θ . . . (1)

-x = cosθ

x = -cosθ

x = cos(π- θ)

π – θ = cos⁻¹x

θ = π – cos⁻¹x  . . . (2)

we know that x ϵ [-1, 1] and $$\pi -\theta \in \left[ 0,\pi \right]\ \ for\ \ all\ \ \theta \in \left[0,\pi \right]$$,

From equation (1) and (2)

$${{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}x\ \ for\ \ all\ \ x\in \left[ -1,1 \right]$$,

Hence proved.