Inverse Trigonometric Functions

Inverse trigonometric functions allow us to work backward: instead of inputting an angle to find a ratio, we input a ratio to find the corresponding angle. The primary inverse functions are $\arcsin(x)$ (or $\sin^{-1}(x)$), $\arccos(x)$ (or $\cos^{-1}(x)$), and $\arctan(x)$ (or $\tan^{-1}(x)$).

Restricted Domains and Ranges

Trigonometric functions are periodic, meaning they repeat their values infinitely. To ensure their inverses are true functions (where each input has exactly one output), we must restrict their domains and ranges.

• $\arcsin(x)$:
  • Domain: $[-1, 1]$
  • Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$ (Quadrants I and IV)
• $\arccos(x)$:
  • Domain: $[-1, 1]$
  • Range: $[0, \pi]$ (Quadrants I and II)
• $\arctan(x)$:
  • Domain: $(-\infty, \infty)$
  • Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$ (Quadrants I and IV)

Evaluating Inverse Functions

When evaluating an expression like $\arcsin(\sin(\theta))$, you must be careful. The answer is only $\theta$ if $\theta$ falls within the restricted range of the inverse function.

Example: Evaluate $\arcsin(\sin(\frac{5\pi}{6}))$

1. First, evaluate the inner function: $\sin(\frac{5\pi}{6}) = \frac{1}{2}$.
2. Now, find $\arcsin(\frac{1}{2})$. We need the angle in the restricted range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ whose sine is $\frac{1}{2}$.
3. The correct angle is $\frac{\pi}{6}$.

Thus, $\arcsin(\sin(\frac{5\pi}{6})) = \frac{\pi}{6}$.

Composite Expressions with Different Functions

Sometimes you need to evaluate expressions like $\tan(\arccos(x))$. The best approach is to model the inner inverse function using a right triangle.

Example: Find the exact value of $\tan(\arccos(\frac{4}{5}))$

1. Let $\theta = \arccos(\frac{4}{5})$. This means $\cos(\theta) = \frac{4}{5}$.
2. In a right triangle, $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$. So, the adjacent side is $4$ and the hypotenuse is $5$.
3. Use the Pythagorean theorem to find the opposite side: $a^2 + b^2 = c^2 \implies \text{Opposite}^2 + 4^2 = 5^2$ $\text{Opposite} = \sqrt{25 - 16} = 3$
4. Now, we want to find $\tan(\theta)$, which is defined as $\frac{\text{Opposite}}{\text{Adjacent}}$.
5. Substituting our triangle's sides, $\tan(\theta) = \frac{3}{4}$.

Therefore, $\tan(\arccos(\frac{4}{5})) = \frac{3}{4}$.