18. Inverse Trigonometric Functions

Adapted from: Classical single-variable calculus (restricted domains for one-to-one trig, inverse definitions, derivative formulas). Rewritten for teaching — not a verbatim textbook dump.

Sine, cosine, and the other trig functions are not one-to-one on their natural domains, so they have no inverses there. Restrict the domain until the graph passes the horizontal-line test; then invert.

---

The inverse sine

On $[-\pi/2,\pi/2]$, $\sin$ is one-to-one (see the sin graph from Ch. 17). Define $\sin^{-1}x$ (also written $\arcsin x$) as the inverse of that restriction.

1. $\sin^{-1}x=y$ iff $\sin y=x$
2. Domain of $\sin^{-1}$: $[-1,1]$ (range of sine)
3. Range of $\sin^{-1}$: $[-\pi/2,\pi/2]$

The graph is the reflection of the restricted sine graph across $y=x$.

Fig. 18-1 — y = arcsin x

Example 18.1 — special values

$\arcsin 0=0$, $\arcsin 1=\pi/2$, $\arcsin(-1)=-\pi/2$, $\arcsin(1/2)=\pi/6$, $\arcsin(\sqrt{2}/2)=\pi/4$, $\arcsin(\sqrt{3}/2)=\pi/3$.

Oddness: $\arcsin(-x)=-\arcsin x$ (because $\sin$ is odd on the restricted interval). So $\arcsin(-1/2)=-\pi/6$.

Book typo

Some printings write $\sin^{-1}(-1/2)=\pi/6$. The correct value is $-\pi/6$.

Check: arcsin

Quiz

Arcsine

1 / 2

arcsin(√3/2) equals…

Pick an answer to see if you are right.

---

Derivative of $\arcsin x$

Let $y=\arcsin x$. Then $\sin y=x$. Implicit differentiation:

$$\cos y\cdot y'=1\implies y'=\frac{1}{\cos y}.$$

$\cos^{2}y=1-\sin^{2}y=1-x^{2}$. On the range $[-\pi/2,\pi/2]$ we have $\cos y\ge 0$, so $\cos y=\sqrt{1-x^{2}}$. Therefore

$$D_{x}(\arcsin x)=\frac{1}{\sqrt{1-x^{2}}}. \tag{18.1}$$

Hands On: Numerical check of $(\arcsin)'$

---

The inverse cosine

Restrict $\cos$ to $[0,\pi]$ (one-to-one, range $[-1,1]$). Define $\arccos x=\cos^{-1}x$:

1. $\arccos x=y$ iff $\cos y=x$
2. Domain $[-1,1]$
3. Range $[0,\pi]$

Reflect the restricted cosine graph across $y=x$.

Fig. 18-2 — y = arccos x

A similar implicit argument gives

$$D_{x}(\arccos x)=-\frac{1}{\sqrt{1-x^{2}}}. \tag{18.2}$$

(Useful check: $\arcsin x+\arccos x=\pi/2$, so the derivatives sum to $0$.)

---

The inverse tangent

Restrict $\tan$ to $(-\pi/2,\pi/2)$ (one-to-one onto $\mathbb{R}$). Define $\arctan x=\tan^{-1}x$:

1. $\arctan x=y$ iff $\tan y=x$
2. Domain $(-\infty,+\infty)$
3. Range $(-\pi/2,\pi/2)$

Example 18.2

$\arctan 0=0$, $\arctan 1=\pi/4$, $\arctan\sqrt{3}=\pi/3$, $\arctan(\sqrt{3}/3)=\pi/6$. Odd: $\arctan(-x)=-\arctan x$.

Graph: reflection of the principal $\tan$ branch; horizontal asymptotes $y=\pm\pi/2$.

Fig. 18-3 — y = arctan x

If $y=\arctan x$, then $\tan y=x$ and $\sec^{2}y\cdot y'=1$, so

$$y'=\frac{1}{\sec^{2}y}=\frac{1}{1+\tan^{2}y}=\frac{1}{1+x^{2}}.$$ $$D_{x}(\arctan x)=\frac{1}{1+x^{2}}. \tag{18.3}$$

Check: arccos / arctan

Quiz

Arccos and arctan

1 / 2

Range of arccos is…

Pick an answer to see if you are right.

---

Inverse cotangent, secant, cosecant

Domains of the inverses are chosen so the derivative formulas stay simple. Conventions for $\mathrm{sec}^{-1}$ and $\mathrm{csc}^{-1}$ vary by textbook — we follow this chapter’s choices.

$\cot^{-1}x$

Restrict $\cot$ to $(0,\pi)$. Domain of $\cot^{-1}$: $\mathbb{R}$; $\cot^{-1}x=y$ iff $\cot y=x$.

$$D_{x}(\cot^{-1}x)=-\frac{1}{1+x^{2}}. \tag{18.4}$$

Fig. 18-4 — cot on (0,π) and cot⁻¹

$\sec^{-1}x$

Restrict $\sec$ to $[0,\pi/2)\cup[\pi,3\pi/2)$. Domain of $\sec^{-1}$: $|x|\ge 1$.

$$D_{x}(\sec^{-1}x)=\frac{1}{x\sqrt{x^{2}-1}}. \tag{18.5}$$

Fig. 18-5 — y = sec⁻¹ x (schematic)

$\csc^{-1}x$

Restrict $\csc$ to $(0,\pi/2]\cup(\pi,3\pi/2]$. Domain $|x|\ge 1$.

$$D_{x}(\csc^{-1}x)=-\frac{1}{x\sqrt{x^{2}-1}}. \tag{18.6}$$

Fig. 18-6 — csc and csc⁻¹ (schematic)

Check: other inverses

Quiz

cot⁻¹ / sec⁻¹ / csc⁻¹

1 / 2

Dₓ(cot⁻¹ x) equals…

Pick an answer to see if you are right.

---

Notation warning

$\sin^{-1}x$ means the inverse function, not $(\sin x)^{-1}=1/\sin x=\csc x$. When ambiguous, write $\arcsin$, $\arccos$, $\arctan$, etc.

---

Key takeaways

• Restrict trig domains to get one-to-one branches; invert to define $\arcsin,\arccos,\arctan,\ldots$
• $(\arcsin)'=1/\sqrt{1-x^{2}}$, $(\arccos)'=-1/\sqrt{1-x^{2}}$, $(\arctan)'=1/(1+x^{2})$
• $(\cot^{-1})'=-1/(1+x^{2})$; $(\sec^{-1})'=1/(x\sqrt{x^{2}-1})$; $(\csc^{-1})'$ flips that sign
• Ranges: $\arcsin\in[-\pi/2,\pi/2]$, $\arccos\in[0,\pi]$, $\arctan\in(-\pi/2,\pi/2)$
• $\sin^{-1}$ ≠ reciprocal — prefer $\arcsin$ when unclear

Further Reading

• 18b. Inverse Trig — Practice — derivatives, values, mural problem
• 19. Rectilinear & Circular Motion — next: v, a, free fall, ω
• 17. Differentiation of Trig Functions — sin′/cos′ and wave graphs
• 17b. Trig Diff — Practice — prior drills
• 16. Review of Trigonometry — unit-circle definitions