Inverse trigonometric functions reverse the action of the regular trig functions. They answer the question: "What angle has this trig value?"

Notation:

• $\arcsin(x)$ or $\sin^{-1}(x)$ is the inverse of sine
• $\arccos(x)$ or $\cos^{-1}(x)$ is the inverse of cosine
• $\arctan(x)$ or $\tan^{-1}(x)$ is the inverse of tangent

Important: $\sin^{-1}(x) \neq \frac{1}{\sin(x)}$. The $-1$ is NOT an exponent!

• $\sin^{-1}(x)$ means inverse sine (arcsin)
• $\frac{1}{\sin(x)} = \csc(x)$ is the reciprocal

Why We Need Restrictions

Trig functions are periodic (repeat), so they're not one-to-one. To have an inverse function, we must restrict the domain.

Restricted Domains (Principal Values)

For $y = \sin(x)$:

• Restrict to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ (Quadrants I and IV)
• This gives all y-values from $-1$ to $1$ exactly once

For $y = \cos(x)$:

• Restrict to $[0, \pi]$ (Quadrants I and II)
• This gives all y-values from $-1$ to $1$ exactly once

For $y = \tan(x)$:

• Restrict to $(-\frac{\pi}{2}, \frac{\pi}{2})$ (Quadrants I and IV)
• This gives all real y-values exactly once

Inverse Sine: $y = \arcsin(x)$

Definition: $y = \arcsin(x)$ means $\sin(y) = x$ where $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

Domain: $[-1, 1]$ (input must be a valid sine value)

Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$ (output is an angle)

Key values:

• $\arcsin(0) = 0$
• $\arcsin(\frac{1}{2}) = \frac{\pi}{6}$
• $\arcsin(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}$
• $\arcsin(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$
• $\arcsin(1) = \frac{\pi}{2}$
• $\arcsin(-1) = -\frac{\pi}{2}$

Inverse Cosine: $y = \arccos(x)$

Definition: $y = \arccos(x)$ means $\cos(y) = x$ where $0 \leq y \leq \pi$

Domain: $[-1, 1]$

Range: $[0, \pi]$

Key values:

• $\arccos(1) = 0$
• $\arccos(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}$
• $\arccos(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}$
• $\arccos(\frac{1}{2}) = \frac{\pi}{3}$
• $\arccos(0) = \frac{\pi}{2}$
• $\arccos(-1) = \pi$

Inverse Tangent: $y = \arctan(x)$

Definition: $y = \arctan(x)$ means $\tan(y) = x$ where $-\frac{\pi}{2} < y < \frac{\pi}{2}$

Domain: All real numbers $(-\infty, \infty)$

Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$

Key values:

• $\arctan(0) = 0$
• $\arctan(1) = \frac{\pi}{4}$
• $\arctan(\sqrt{3}) = \frac{\pi}{3}$
• $\arctan(-1) = -\frac{\pi}{4}$

Asymptotes:

• $\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}$
• $\lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}$

Composition Properties

Inverse function property:

For values in the appropriate domains:

• $\sin(\arcsin(x)) = x$ for $x \in [-1, 1]$
• $\arcsin(\sin(x)) = x$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$

Similarly for cosine and tangent.

Warning: $\arcsin(\sin(x))$ does NOT always equal $x$!

Example: $\arcsin(\sin(\frac{3\pi}{4})) = \frac{\pi}{4}$ (not $\frac{3\pi}{4}$)

Why? Because $\frac{3\pi}{4}$ is outside the range of arcsin.

Using Inverse Trig to Solve Equations

Example: Solve $\sin(x) = 0.7$ for $0 \leq x < 2\pi$

Step 1: Find the reference angle $x_{ref} = \arcsin(0.7) \approx 0.775 \text{ radians}$

Step 2: Determine quadrants (sine is positive in I and II)

• Quadrant I: $x = 0.775$
• Quadrant II: $x = \pi - 0.775 \approx 2.366$

Solutions: $x \approx 0.775, 2.366$

Graphs of Inverse Trig Functions

All inverse trig functions are reflections of the restricted trig functions over the line $y = x$.

Graph characteristics:

$y = \arcsin(x)$:

• Domain: $[-1, 1]$, Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$
• Increasing function
• Passes through origin

$y = \arccos(x)$:

• Domain: $[-1, 1]$, Range: $[0, \pi]$
• Decreasing function
• $y$-intercept at $(0, \frac{\pi}{2})$

$y = \arctan(x)$:

• Domain: $(-\infty, \infty)$, Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$
• Increasing function
• Horizontal asymptotes at $y = \pm\frac{\pi}{2}$
• Passes through origin

Other Inverse Trig Functions

• $y = \arccsc(x) = \arcsin(\frac{1}{x})$ for $|x| \geq 1$
• $y = \arcsec(x) = \arccos(\frac{1}{x})$ for $\mathrm{\mid }x\mathrm{\mid }$
• $y = \arccot(x) = \arctan(\frac{1}{x})$ (with adjustments)

These are less commonly used but follow similar principles.

From the unit circle:

$\sin(\frac{\pi}{6}) = \frac{1}{2}$

And $\frac{\pi}{6}$ is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ ✓

Answer: $\arcsin(\frac{1}{2}) = \frac{\pi}{6}$

Verification: $\sin(\frac{\pi}{6}) = \frac{1}{2}$ ✓

Note: Even though $\sin(\frac{5\pi}{6}) = \frac{1}{2}$ as well, we don't choose $\frac{5\pi}{6}$ because it's outside the range of arcsin.