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.

We need to find: $\sin(\theta)$

Method: Use Pythagorean identity

$\sin^2(\theta) + \cos^2(\theta) = 1$

Substitute $\cos(\theta) = \frac{3}{5}$: $\sin^2(\theta) + (\frac{3}{5})^2 = 1$ $\sin^2(\theta) + \frac{9}{25} = 1$ $\sin^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25}$ $\sin(\theta) = \pm\frac{4}{5}$

Determine the sign:

Since $\theta = \arccos(\frac{3}{5})$ and the range of arccos is $[0, \pi]$, $\theta$ is in Quadrant I or II.

In both quadrants, sine is positive.

Answer: $\sin(\arccos(\frac{3}{5})) = \frac{4}{5}$

Alternative method (right triangle):

If $\cos(\theta) = \frac{3}{5} = \frac{\text{adjacent}}{\text{hypotenuse}}$:

• Adjacent = 3
• Hypotenuse = 5
• Opposite = $\sqrt{5^2 - 3^2} = \sqrt{16} = 4$

Therefore: $\sin(\theta) = \frac{4}{5}$ ✓

Step 1: Isolate one arctan $\arctan(x) = \frac{\pi}{4} - \arctan(2x)$

Step 2: Take tangent of both sides $x = \tan(\frac{\pi}{4} - \arctan(2x))$

Step 3: Use tangent difference formula

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Here: $A = \frac{\pi}{4}$, $B = \arctan(2x)$

$x = \frac{\tan(\frac{\pi}{4}) - \tan(\arctan(2x))}{1 + \tan(\frac{\pi}{4})\tan(\arctan(2x))}$

Step 4: Simplify

Since $\tan(\frac{\pi}{4}) = 1$ and $\tan(\arctan(2x)) = 2x$:

$x = \frac{1 - 2x}{1 + (1)(2x)} = \frac{1 - 2x}{1 + 2x}$

Step 5: Solve for $x$

$x(1 + 2x) = 1 - 2x$ $x + 2x^2 = 1 - 2x$ $2x^2 + x + 2x - 1 = 0$ $2x^2 + 3x - 1 = 0$

Step 6: Use quadratic formula

$x = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$

Step 7: Check which solution is valid

$x_1 = \frac{-3 + \sqrt{17}}{4} \approx 0.281$

$x_2 = \frac{-3 - \sqrt{17}}{4} \approx -1.781$

Both are in the domain of arctan, so check by substitution:

For $x = \frac{-3 + \sqrt{17}}{4}$: Calculate $\arctan(x) + \arctan(2x)$ numerically to verify it equals $\frac{\pi}{4} \approx 0.785$

Answer: $x = \frac{-3 + \sqrt{17}}{4}$

(The negative solution may or may not work depending on domain restrictions)