Math Core

So far you have gone from an angle to a number: $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$ . Many real problems run the other way. A ramp rises $1$ m along $4$ m of sloped surface; what angle does it make? To answer, you need functions that take a ratio and return an angle. Those are the inverse trigonometric functions, and the only subtlety is choosing which angle they return.

The problem: trig functions are not one-to-one

A function has an inverse only if it is one-to-one (it passes the horizontal line test). Sine fails badly: the line $y = \dfrac{1}{2}$ crosses $y = \sin x$ at $\dfrac{\pi}{6}$ , $\dfrac{5\pi}{6}$ , $\dfrac{13\pi}{6}$ and infinitely many other places. So "the angle whose sine is $\dfrac{1}{2}$ " is ambiguous.

The fix is the same one you use for $y = x^2$ , where you restrict to $x \ge 0$ to define $\sqrt{x}$ . Restrict each trig function to an interval where it is one-to-one, still hits every output, and contains the first-quadrant angles.

Sine on $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ : it increases from $-1$ to $1$ .
Cosine on $[0, \pi]$ : it decreases from $1$ to $-1$ .
Tangent on $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ : it increases through all real numbers.

The three inverse functions

Definition

Inverse sine, cosine and tangent

\begin{aligned} y = \arcsin x \quad &\text{means} \quad \sin y = x \text{ and } -\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}, && \text{domain } [-1, 1]\\ y = \arccos x \quad &\text{means} \quad \cos y = x \text{ and } 0 \le y \le \pi, && \text{domain } [-1, 1]\\ y = \arctan x \quad &\text{means} \quad \tan y = x \text{ and } -\tfrac{\pi}{2} \lt y \lt \tfrac{\pi}{2}, && \text{domain } (-\infty, \infty) \end{aligned}

They are also written $\sin^{-1} x$ , $\cos^{-1} x$ and $\tan^{-1} x$ .

The restricted intervals are now the ranges of the inverse functions. On the unit circle, arcsin and arctan always return an angle on the right half (Quadrant I or IV, as a negative angle), while arccos returns an angle on the top half (Quadrant I or II).

Common mistake

$\sin^{-1} x$ does not mean $\dfrac{1}{\sin x}$ . The $-1$ here means inverse function. The reciprocal is $(\sin x)^{-1} = \csc x$ . Also, $\arcsin\left(-\dfrac{1}{2}\right)$ is $-\dfrac{\pi}{6}$ , not $\dfrac{11\pi}{6}$ : the answer must lie in the range.

Graphically, each inverse is the reflection of the restricted function across $y = x$ . The arcsine graph runs from $\left(-1, -\dfrac{\pi}{2}\right)$ to $\left(1, \dfrac{\pi}{2}\right)$ ; arccosine runs from $(-1, \pi)$ down to $(1, 0)$ ; arctangent has horizontal asymptotes $y = \pm\dfrac{\pi}{2}$ .

y = arcsin x, y = arccos x and y = arctan x. The dashed lines y = ±π/2 are asymptotes of arctan.Open in grapher →

Evaluating exactly

To evaluate $\arcsin a$ , ask: which angle in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ has sine $a$ ? Use the reference angle from the special-values table, then pick the sign or quadrant allowed by the range.

Worked example: Three exact values

Find $\arcsin\left(-\dfrac{\sqrt{2}}{2}\right)$ , $\arccos\left(-\dfrac{\sqrt{3}}{2}\right)$ and $\arctan\sqrt{3}$ .

Solution. The reference angle for $\dfrac{\sqrt{2}}{2}$ is $\dfrac{\pi}{4}$ . Arcsin returns a Quadrant IV angle for negative inputs, so $\arcsin\left(-\dfrac{\sqrt{2}}{2}\right) = -\dfrac{\pi}{4}$ .

The reference angle for $\dfrac{\sqrt{3}}{2}$ (as a cosine) is $\dfrac{\pi}{6}$ . Arccos returns a Quadrant II angle for negative inputs, so $\arccos\left(-\dfrac{\sqrt{3}}{2}\right) = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ .

$\tan\dfrac{\pi}{3} = \sqrt{3}$ and $\dfrac{\pi}{3}$ is in the range, so $\arctan\sqrt{3} = \dfrac{\pi}{3}$ .

Tip

For negative inputs there are quick rules: $\arcsin(-x) = -\arcsin x$ and $\arctan(-x) = -\arctan x$ (both are odd functions), but $\arccos(-x) = \pi - \arccos x$ .

Compositions

An inverse undoes its function, but only in one direction without conditions.

Cancellation rules

\sin(\arcsin x) = x \ \text{ for } -1 \le x \le 1, \qquad \arcsin(\sin x) = x \ \text{ only for } -\tfrac{\pi}{2} \le x \le \tfrac{\pi}{2}.

The same pattern holds for cosine (with $0 \le x \le \pi$ ) and tangent (with $-\tfrac{\pi}{2} \lt x \lt \tfrac{\pi}{2}$ ). Outside those intervals, evaluate the inside first.

Worked example: When the inverse does not cancel

Find $\arccos\left(\cos\dfrac{5\pi}{3}\right)$ and $\arcsin\left(\sin\dfrac{3\pi}{4}\right)$ .

Solution. $\dfrac{5\pi}{3}$ is not in $[0, \pi]$ , so work inside out: $\cos\dfrac{5\pi}{3} = \dfrac{1}{2}$ , and $\arccos\dfrac{1}{2} = \dfrac{\pi}{3}$ .

$\dfrac{3\pi}{4}$ is not in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ : $\sin\dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}$ , and $\arcsin\dfrac{\sqrt{2}}{2} = \dfrac{\pi}{4}$ .

When a trig function is applied to an inverse of a different trig function, draw a triangle. Call the inner angle $\theta$ , label two sides from its definition, and find the third with the Pythagorean theorem. The range tells you the sign.

Worked example: The triangle method

Find $\tan\left(\arccos\left(-\dfrac{3}{5}\right)\right)$ , then write $\cos(\arcsin x)$ as an algebraic expression.

Solution. Let $\theta = \arccos\left(-\dfrac{3}{5}\right)$ . Then $\cos\theta = -\dfrac{3}{5}$ with $\theta$ in $[0, \pi]$ ; since the cosine is negative, $\theta$ is in Quadrant II. A reference triangle with adjacent side $3$ and hypotenuse $5$ has opposite side $4$ . In Quadrant II, $x = -3$ and $y = 4$ , so $\tan\theta = -\dfrac{4}{3}$ .

For the second part, let $\theta = \arcsin x$ , so $\sin\theta = x$ with $\theta$ in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ . On that interval cosine is never negative, so

\cos(\arcsin x) = \sqrt{1 - \sin^2\theta} = \sqrt{1 - x^2}.

Solving equations with a calculator

A calculator's inverse keys answer the ramp question from the introduction directly: the angle satisfies $\sin\theta = \dfrac{1}{4}$ , so $\theta = \arcsin 0.25 \approx 0.253$ radians, or about $14.5^\circ$ .

For equations, inverse functions give one solution of an equation like $\sin x = 0.3$ . The unit circle gives the rest. Since $\arcsin 0.3 \approx 0.305$ is in Quadrant I, the other solution in $[0, 2\pi)$ has the same reference angle in Quadrant II: $\pi - 0.305 \approx 2.837$ . Add $2\pi k$ to either for all solutions. The next unit develops this fully.