Inverse Trigonometric Functions

You already know from Trigonometric Functions that $\sin$, $\cos$, and $\tan$ each map an angle to a ratio. The natural inverse question is: given a ratio, which angle produced it? Answering it requires inverse trigonometric functions — but there is a catch. Because sine and cosine are periodic, they repeat every $2\pi$ and are therefore not injective on all of $\mathbb{R}$; a function with repeated outputs cannot be inverted globally. The standard remedy is to restrict each function to a smaller domain where it is strictly monotone and hence injective, and then invert it there.

Arcsine

On the interval $\left[-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right]$, the sine function is strictly increasing from $-1$ to $1$ — every value in $[-1, 1]$ is achieved exactly once. This restriction is injective, so it has an inverse.

Arcsine is defined as

$$\arcsin : [-1,\ 1] \to \left[-\frac{\pi}{2},\ \frac{\pi}{2}\right],$$

where $\arcsin x$ is the unique angle $\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ satisfying $\sin\theta = x$.

The inverse relationship yields two composition identities:

$$\sin(\arcsin x) = x \quad \text{for all } x \in [-1,\ 1],$$ $$\arcsin(\sin\theta) = \theta \quad \text{for all } \theta \in \left[-\frac{\pi}{2},\ \frac{\pi}{2}\right].$$

The second identity holds only on the restricted interval. Outside it, $\arcsin(\sin\theta)$ gives the representative of $\theta$ inside $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, not $\theta$ itself.

Derivative of arcsine

Let $\theta = \arcsin x$, so $x = \sin\theta$ with $\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Differentiating both sides with respect to $x$:

$$1 = \cos\theta \cdot \frac{d\theta}{dx}.$$

On $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, cosine is non-negative, so $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - x^2}$. Solving for $\dfrac{d\theta}{dx}$:

$$\arcsin'(x) = \frac{1}{\sqrt{1 - x^2}}, \qquad x \in (-1,\ 1).$$

The derivative is undefined at the endpoints $x = \pm 1$, where the graph of $\arcsin$ has a vertical tangent.

Arccosine

On the interval $[0, \pi]$, the cosine function is strictly decreasing from $1$ to $-1$. The restriction is injective.

Arccosine is defined as

$$\arccos : [-1,\ 1] \to [0,\ \pi],$$

where $\arccos x$ is the unique angle $\theta \in [0, \pi]$ satisfying $\cos\theta = x$.

Derivative of arccosine

Let $\theta = \arccos x$, so $x = \cos\theta$. Differentiating with respect to $x$ gives $1 = -\sin\theta \cdot \dfrac{d\theta}{dx}$. On $[0, \pi]$, sine is non-negative, so $\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - x^2}$. Solving:

$$\arccos'(x) = -\frac{1}{\sqrt{1 - x^2}}, \qquad x \in (-1,\ 1).$$

The negative sign reflects the fact that $\arccos$ is strictly decreasing.

The arcsin–arccos identity

For all $x \in [-1, 1]$:

$$\arcsin x + \arccos x = \frac{\pi}{2}.$$

Proof. Let $\alpha = \arcsin x$, so $\sin\alpha = x$ and $\alpha \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Then

$$\cos\!\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha = x,$$

and $\dfrac{\pi}{2} - \alpha \in [0, \pi]$. By the uniqueness in the definition of $\arccos$, it follows that $\arccos x = \dfrac{\pi}{2} - \alpha$, which rearranges to the claimed identity. $\square$

This identity lets you convert between $\arcsin$ and $\arccos$ without any recomputation, and it also explains why their derivatives are negatives of each other.

Arctangent

On the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, the tangent function is strictly increasing and takes every real value exactly once. (The endpoints are excluded because $\tan$ is undefined there.)

Arctangent is defined as

$$\arctan : \mathbb{R} \to \left(-\frac{\pi}{2},\ \frac{\pi}{2}\right),$$

where $\arctan x$ is the unique angle $\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ satisfying $\tan\theta = x$.

Unlike $\arcsin$ and $\arccos$, the domain of $\arctan$ is all of $\mathbb{R}$. The output is bounded, however, so the function has two horizontal asymptotes:

$$\lim_{x \to +\infty} \arctan x = \frac{\pi}{2}, \qquad \lim_{x \to -\infty} \arctan x = -\frac{\pi}{2}.$$

Derivative of arctangent

Let $\theta = \arctan x$, so $\tan\theta = x$. Differentiating with respect to $x$:

$$\sec^2\!\theta \cdot \frac{d\theta}{dx} = 1.$$

Using the Pythagorean identity $\sec^2\theta = 1 + \tan^2\theta = 1 + x^2$:

$$\arctan'(x) = \frac{1}{1 + x^2}, \qquad x \in \mathbb{R}.$$

This is defined for all real $x$ and is always positive, confirming that $\arctan$ is strictly increasing on its entire domain.

Other inverse trigonometric functions

The remaining three inverse trigonometric functions are obtained by restricting $\cot$, $\sec$, and $\csc$ to standard intervals of injectivity.

Arccotangent is the inverse of $\cot$ restricted to $(0, \pi)$:

$$\operatorname{arccot} : \mathbb{R} \to (0,\ \pi).$$

Arcsecant is the inverse of $\sec$ restricted to $[0, \pi] \setminus \left\{\dfrac{\pi}{2}\right\}$:

$$\operatorname{arcsec} : (-\infty, -1] \cup [1, +\infty) \to [0,\ \pi] \setminus \left\{\frac{\pi}{2}\right\}.$$

Arccosecant is the inverse of $\csc$ restricted to $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \setminus \{0\}$:

$$\operatorname{arccsc} : (-\infty, -1] \cup [1, +\infty) \to \left[-\frac{\pi}{2},\ \frac{\pi}{2}\right] \setminus \{0\}.$$

These three arise less often in practice. When you encounter them, they can usually be rewritten in terms of $\arcsin$, $\arccos$, or $\arctan$.

Table of derivatives

The derivatives of all six inverse trigonometric functions, collected for reference:

Function	Derivative	Domain of derivative
$\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$	$(-1,\ 1)$
$\arccos x$	$-\dfrac{1}{\sqrt{1-x^2}}$	$(-1,\ 1)$
$\arctan x$	$\dfrac{1}{1+x^2}$	$\mathbb{R}$
$\operatorname{arccot} x$	$-\dfrac{1}{1+x^2}$	$\mathbb{R}$
$\operatorname{arcsec} x$	$\dfrac{1}{\lvert x \rvert\sqrt{x^2-1}}$	$\lvert x\rvert > 1$
$\operatorname{arccsc} x$	$-\dfrac{1}{\lvert x \rvert \sqrt{x^2-1}}$	$\lvert x\rvert > 1$

Notice the pattern: in each pair, $\arcsin$ and $\arccos$, $\arctan$ and $\operatorname{arccot}$, $\operatorname{arcsec}$ and $\operatorname{arccsc}$, the derivatives are negatives of each other. This reflects the complementary identities $\arcsin x + \arccos x = \dfrac{\pi}{2}$ and $\arctan x + \operatorname{arccot} x = \dfrac{\pi}{2}$.

Summary

• Because $\sin$, $\cos$, and $\tan$ are not injective on all of $\mathbb{R}$, their inverses require restricting to a carefully chosen interval of strict monotonicity.
• Arcsine: $\arcsin : [-1, 1] \to \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, with derivative $\dfrac{1}{\sqrt{1-x^2}}$ on $(-1, 1)$.
• Arccosine: $\arccos : [-1, 1] \to [0, \pi]$, with derivative $-\dfrac{1}{\sqrt{1-x^2}}$ on $(-1, 1)$; satisfies the identity $\arcsin x + \arccos x = \dfrac{\pi}{2}$.
• Arctangent: $\arctan : \mathbb{R} \to \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, with derivative $\dfrac{1}{1+x^2}$ on $\mathbb{R}$; has horizontal asymptotes $\pm\dfrac{\pi}{2}$.
• The three additional inverses $\operatorname{arccot}$, $\operatorname{arcsec}$, $\operatorname{arccsc}$ follow the same pattern with analogous restricted domains.
• All six derivatives are derived via the inverse function theorem (differentiate the defining identity $f(\theta) = x$ with respect to $x$) and result in algebraic expressions, even though the original trigonometric functions are transcendental.