8.3 Inverse Trigonometric Functions

Inverse Trigonometric Functions

Regular trig functions take an angle and give you a ratio. Inverse trig functions do the opposite: you feed in a ratio, and they return an angle. This is essential whenever you need to solve for an unknown angle in an equation like $\sin \theta = 0.5$.

Because trig functions repeat their values over and over (they're periodic), we have to restrict their domains before creating inverses. Otherwise, a single input like $0.5$ would map to infinitely many angles, and that's not a function. The restricted domains give each inverse trig function a specific range of output angles.

Inverse Trigonometric Functions

Inverse trig functions are written with a superscript $-1$ or with the "arc" prefix: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ (or arcsin, arccos, arctan). The $-1$ is notation for the inverse, not an exponent.

Each inverse function has its own domain (what you can put in) and range (what angles come out):

• Inverse sine ($\sin^{-1}$ or arcsin): domain $[-1, 1]$, range $[-\frac{\pi}{2}, \frac{\pi}{2}]$
• Inverse cosine ($\cos^{-1}$ or arccos): domain $[-1, 1]$, range $[0, \pi]$
• Inverse tangent ($\tan^{-1}$ or arctan): domain $(-\infty, \infty)$, range $(-\frac{\pi}{2}, \frac{\pi}{2})$

Notice that arcsin and arctan use the same interval but arcsin includes the endpoints (brackets) while arctan excludes them (parentheses). Arccos covers a completely different range: from $0$ to $\pi$.

The core idea is straightforward:

• If $\sin \theta = x$, then $\theta = \sin^{-1}(x)$
• If $\cos \theta = x$, then $\theta = \cos^{-1}(x)$
• If $\tan \theta = x$, then $\theta = \tan^{-1}(x)$

Graphically, each inverse trig function is the reflection of the (restricted) original function over the line $y = x$.

Exact Values of Inverse Trigonometry

You can find exact values by thinking backward through the unit circle. Ask yourself: which angle in the allowed range produces this ratio?

The angles you'll use most often are $\frac{\pi}{6}$ (30°), $\frac{\pi}{4}$ (45°), and $\frac{\pi}{3}$ (60°), along with $0$ and $\frac{\pi}{2}$.

Examples:

• $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ because $\sin(\frac{\pi}{6}) = \frac{1}{2}$, and $\frac{\pi}{6}$ is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$
• $\cos^{-1}(-\frac{\sqrt{2}}{2}) = \frac{3\pi}{4}$ because $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$, and $\frac{3\pi}{4}$ is in $[0, \pi]$
• $\tan^{-1}(1) = \frac{\pi}{4}$ because $\tan(\frac{\pi}{4}) = 1$, and $\frac{\pi}{4}$ is in $(-\frac{\pi}{2}, \frac{\pi}{2})$

A common mistake is giving an angle outside the allowed range. For instance, $\sin^{-1}(\frac{1}{2}) \neq \frac{5\pi}{6}$ even though $\sin(\frac{5\pi}{6}) = \frac{1}{2}$, because $\frac{5\pi}{6}$ falls outside $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Always check that your answer is in the correct range.

Answers are typically given in radians unless a problem specifically asks for degrees.

Technology in Inverse Trigonometry

Scientific and graphing calculators have dedicated inverse trig buttons, usually labeled $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ (or arcsin, arccos, arctan). On most calculators, you access them by pressing a "2nd" or "shift" key first.

To evaluate $\sin^{-1}(0.8)$:

1. Make sure your calculator is set to the mode you want (radian or degree).
2. Press the "2nd" or "shift" key, then press "sin."
3. Enter $0.8$ and press enter/equals.
4. You'll get approximately $0.9273$ radians (or about $53.13°$ in degree mode).

Calculators default to radians on many models, so double-check your mode setting before you start. Getting an answer of $0.9273$ when you expected something near $53$ (or vice versa) almost always means you're in the wrong mode.

Composite Inverse Trigonometric Functions

A composite expression like $\sin(\cos^{-1}(x))$ nests one function inside another. To evaluate these, always work from the inside out.

Steps:

1. Evaluate the inner inverse trig function to get an angle.
2. Apply the outer trig function to that angle.
3. Simplify if possible.

Worked example: Evaluate $\sin(\cos^{-1}(\frac{\sqrt{3}}{2}))$

1. Start inside: $\cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}$ because $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\frac{\pi}{6}$ is in $[0, \pi]$.
2. Apply the outer function: $\sin(\frac{\pi}{6}) = \frac{1}{2}$.
3. Final answer: $\frac{1}{2}$.

For values that don't correspond to standard unit circle angles, you can sketch a right triangle. If the inner function is $\cos^{-1}(\frac{3}{5})$, for example, draw a right triangle where the adjacent side is $3$ and the hypotenuse is $5$. The Pythagorean theorem gives the opposite side as $4$, so $\sin(\cos^{-1}(\frac{3}{5})) = \frac{4}{5}$.

Trigonometric Ratios and the Unit Circle

The unit circle is a circle with radius $1$ centered at the origin. For any angle $\theta$, the point on the unit circle is $(\cos \theta, \sin \theta)$. That means the $x$-coordinate gives cosine and the $y$-coordinate gives sine. Tangent is the ratio $\frac{\sin \theta}{\cos \theta}$.

Inverse trig functions reverse this process: given a coordinate (ratio), you find the corresponding angle. Visualizing the unit circle helps you confirm whether your answer falls in the correct range and whether the sign (positive or negative) makes sense for that quadrant.