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.