A
Composition: f⁻¹(f(x)) = x
Applying the inverse after the original function returns the argument
Key Idea
These hold only when $x$ lies in the principal value branch of the inverse function. Outside this range, a correction may be needed.
$\sin^{-1}(\sin x) = x$
x ∈ [−π/2, π/2]
$\cos^{-1}(\cos x) = x$
x ∈ [0, π]
$\tan^{-1}(\tan x) = x$
x ∈ (−π/2, π/2)
$\cot^{-1}(\cot x) = x$
x ∈ (0, π)
$\sec^{-1}(\sec x) = x$
x ∈ [0, π], x ≠ π/2
$\csc^{-1}(\csc x) = x$
x ∈ [−π/2, π/2], x ≠ 0
B
Composition: f(f⁻¹(x)) = x
Applying the original function after the inverse gives back the argument
$\sin(\sin^{-1} x) = x$
x ∈ [−1, 1]
$\cos(\cos^{-1} x) = x$
x ∈ [−1, 1]
$\tan(\tan^{-1} x) = x$
x ∈ ℝ
$\cot(\cot^{-1} x) = x$
x ∈ ℝ
$\sec(\sec^{-1} x) = x$
x ∈ (−∞,−1] ∪ [1,∞)
$\csc(\csc^{-1} x) = x$
x ∈ (−∞,−1] ∪ [1,∞)
C
Reciprocal Relations
Inverse of a reciprocal trig function equals the co-function's inverse
Remember: $\sin^{-1}\!\left(\tfrac{1}{x}\right) \neq \dfrac{1}{\sin^{-1}x}$. These are two entirely different things. The properties below express reciprocal arguments in terms of different inverse functions.
$$\sin^{-1}\!\left(\frac{1}{x}\right) = \csc^{-1} x, \quad |x| \geq 1$$
$$\cos^{-1}\!\left(\frac{1}{x}\right) = \sec^{-1} x, \quad |x| \geq 1$$
$$\tan^{-1}\!\left(\frac{1}{x}\right) = \begin{cases} \cot^{-1} x & \text{if } x > 0 \\[6pt] -\pi + \cot^{-1} x & \text{if } x < 0 \end{cases}$$
✦
Quick Reference Summary
All Set I properties at a glance
| Identity | Condition / Domain | Group |
|---|---|---|
| $\sin^{-1}(\sin x)=x$ | $x\in[-\pi/2,\,\pi/2]$ | A |
| $\cos^{-1}(\cos x)=x$ | $x\in[0,\pi]$ | A |
| $\tan^{-1}(\tan x)=x$ | $x\in(-\pi/2,\pi/2)$ | A |
| $\sin(\sin^{-1}x)=x$ | $x\in[-1,1]$ | B |
| $\cos(\cos^{-1}x)=x$ | $x\in[-1,1]$ | B |
| $\tan(\tan^{-1}x)=x$ | $x\in\mathbb{R}$ | B |
| $\sin^{-1}(1/x)=\csc^{-1}x$ | $|x|\geq 1$ | C |
| $\cos^{-1}(1/x)=\sec^{-1}x$ | $|x|\geq 1$ | C |
| $\tan^{-1}(1/x)=\cot^{-1}x$ | $x>0$ | C |
| $\tan^{-1}(1/x)=-\pi+\cot^{-1}x$ | $x<0$ | C |