Solving Trig Equations with Inverse Functions

Recap: Previously, we solved trigonometric equations like $\sin(x) = \frac{1}{2}$ using the unit circle to find special angles (e.g., $\frac{\pi}{6}$ , $\frac{5\pi}{6}$ ).
Problem: What if we encounter equations like $\sin(x) = \frac{1}{3}$ , where $\frac{1}{3}$ isn't a standard value on the unit circle?
Solution: Introduce inverse trigonometric functions to solve for angles when the sine, cosine, or tangent is not a standard value.

General Concept: If $f$ is a function, its inverse function, denoted as $f^{-1}$ , "undoes" the effect of $f$ .
Property: $f^{-1}(f(a)) = a$ for any number $a$ in the domain of $f$ .
Application to Trigonometry -- if $\sin(x) = \frac{1}{3}$ , then $x = \sin^{-1}(\frac{1}{3})$ (inverse sine of $\frac{1}{3}$ ).

Sine Function: The sine function, $y = \sin(x)$ , is a wave that oscillates between -1 and 1.
Inverse Function as Reflection: The inverse function is a reflection of the original function across the line $y = x$ .
Issue: Reflecting the entire sine curve across $y = x$ does not result in a function because it fails the vertical line test; a vertical line intersects the reflected curve multiple times.
Points on Sine Curve and Their Reflections: For example, the point $\pi, 0$ on the sine curve becomes $0, \pi$ on the reflected curve.

Solution: To obtain a valid inverse function, we restrict the domain of the original sine function.
Restricted Domain: The domain of the sine function is limited to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ .
Sine Function in Restricted Domain: In this interval, the sine function ranges from -1 to 1 without repeating y-values.
Key Points: The sine function passes through the points $(-\frac{\pi}{2}, -1)$ , $(0, 0)$ , and $(\frac{\pi}{2}, 1)$ in this restricted domain.

Reflection of Restricted Sine Function: Reflecting the restricted sine function across $y = x$ yields the inverse sine function.
Domain and Range: The inverse sine function, denoted as $\sin^{-1}(x)$ or arcsin(x), has a domain of [-1, 1] and a range of $[-\frac{\pi}{2}, \frac{\pi}{2}]$ .
Graph of Inverse Sine: The graph starts at $(-1, -\frac{\pi}{2})$ and ends at $(1, \frac{\pi}{2})$ .
Calculator Usage: Calculators only show the principal values of the inverse sine function due to the domain restriction.

Example: To solve $\sin(x) = \frac{1}{3}$ , take the inverse sine of both sides: $x = \sin^{-1}(\frac{1}{3})$ .
Calculator Result: Using a calculator, $\sin^{-1}(\frac{1}{3}) \approx 0.3398$ radians.
Interpretation: This is the angle in radians whose sine is $\frac{1}{3}$ .
Multiple Solutions: Infinitely many angles have a sine of $\frac{1}{3}$ , but the inverse sine function only returns the angle within the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ .

Inverse Cosine: The cosine function requires a different domain restriction to have an inverse.
- Restricted Domain: The domain of the cosine function it's limited to $[0, \pi]$ .
- Range of Inverse Cosine: The range of the inverse cosine function is $[0, \pi]$ .
Inverse Tangent: The tangent function has vertical asymptotes and a range of $(-\infty, \infty)$ .
- Appropriate Interval: $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is the interval to determine the inverse tangent function uniquely.
- Range of Inverse Tangent: The range of the inverse tangent function is $(-\frac{\pi}{2}, \frac{\pi}{2})$ .

Equation Solutions: When solving trigonometric equations, remember that inverse trigonometric functions only provide one solution within a specific interval.
Multiple Solutions: Equations may have multiple solutions, but inverse functions only give the principal value within their defined range.