Math Core

What happens when you apply a trig function and its inverse one after the other? Sometimes they cancel perfectly, and sometimes they don't, because of the restricted ranges. Understanding compositions like $\sin(\arcsin x)$ , $\arcsin(\sin x)$ and $\sin(\arccos x)$ is essential for simplifying expressions in calculus and for checking answers to trig equations.

Trig function of its own inverse

Start with $\sin(\arcsin x)$ . By definition, $\arcsin x$ is an angle whose sine is $x$ . Taking the sine of that angle gives back $x$ :

\sin\left(\arcsin 0.4\right) = 0.4, \qquad \cos\left(\arccos\left(-\frac{2}{7}\right)\right) = -\frac{2}{7}, \qquad \tan(\arctan 25) = 25.

The only catch is the domain. $\arcsin x$ exists only for $-1 \le x \le 1$ , so $\sin(\arcsin 3)$ is undefined, not $3$ .

Inverse of a trig function

Now reverse the order: $\arcsin(\sin\theta)$ . Here the output of arcsine must land in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ . If $\theta$ is already in that interval, you get $\theta$ back. If not, you get a different angle with the same sine.

For example, $\arcsin\left(\sin\dfrac{\pi}{5}\right) = \dfrac{\pi}{5}$ , but

\arcsin\left(\sin\frac{3\pi}{4}\right) = \arcsin\frac{\sqrt{2}}{2} = \frac{\pi}{4}, \quad\text{not } \frac{3\pi}{4}.

When compositions cancel

Function of inverse (inside out: inverse first). These cancel whenever the expression is defined:

\sin(\arcsin x) = x \ \text{ and } \ \cos(\arccos x) = x \ \text{ for } -1 \le x \le 1, \qquad \tan(\arctan x) = x \ \text{ for all } x.

Inverse of function (trig function first). These cancel only when the angle is in the range of the inverse:

\begin{aligned} \arcsin(\sin\theta) &= \theta \quad\text{only if } -\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2} \\ \arccos(\cos\theta) &= \theta \quad\text{only if } 0 \le \theta \le \pi \\ \arctan(\tan\theta) &= \theta \quad\text{only if } -\tfrac{\pi}{2} < \theta < \tfrac{\pi}{2} \end{aligned}

When the angle is outside the range, work from the inside out: evaluate the trig function first, then find the inverse of that number the usual way.

Worked example: Angles outside the range

Find the exact value of each.

$\arcsin\left(\sin\dfrac{5\pi}{6}\right)$
$\arccos\left(\cos\left(-\dfrac{\pi}{3}\right)\right)$
$\arctan\left(\tan\dfrac{3\pi}{4}\right)$

Solutions.

$\dfrac{5\pi}{6}$ is not in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ , so don't cancel. First $\sin\dfrac{5\pi}{6} = \dfrac{1}{2}$ , and then $\arcsin\dfrac{1}{2} = \dfrac{\pi}{6}$ .
$-\dfrac{\pi}{3}$ is not in $[0, \pi]$ . First $\cos\left(-\dfrac{\pi}{3}\right) = \dfrac{1}{2}$ , and then $\arccos\dfrac{1}{2} = \dfrac{\pi}{3}$ .
$\dfrac{3\pi}{4}$ is not in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ . First $\tan\dfrac{3\pi}{4} = -1$ , and then $\arctan(-1) = -\dfrac{\pi}{4}$ .

In each case the answer is an angle in the range with the same trig value as the original angle.

Common mistake

Don't cancel $\arcsin(\sin\theta)$ or $\arccos(\cos\theta)$ on sight. Check the angle against the range first. $\arccos(\cos 2\pi)$ is $\arccos 1 = 0$ , not $2\pi$ , because an arccosine can never be larger than $\pi$ .

Mixing different functions: draw a triangle

Now for a composition like $\sin\left(\arccos\dfrac{3}{5}\right)$ . The functions are different, so nothing cancels. The trick is to name the angle and draw it.

Let $\theta = \arccos\dfrac{3}{5}$ . This says $\cos\theta = \dfrac{3}{5}$ , with $\theta$ in $[0, \pi]$ . Since the cosine is positive, $\theta$ is in Quadrant I, so you can picture it in a right triangle with adjacent side $3$ and hypotenuse $5$ .

If cos θ = 3/5, draw a right triangle with adjacent side 3 and hypotenuse 5. The Pythagorean theorem gives the opposite side, 4.

The Pythagorean theorem gives the third side: $\sqrt{5^2 - 3^2} = \sqrt{16} = 4$ . Now read off any ratio you like:

\sin\left(\arccos\frac{3}{5}\right) = \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}.

The triangle method

To evaluate $\text{trig}(\text{inverse trig}(a))$ :

Let $\theta$ equal the inner expression, and write what it means (for example, $\sin\theta = a$ ).
Use the range of the inverse function and the sign of $a$ to decide which quadrant $\theta$ is in.
Draw a right triangle for the reference angle and find the missing side with the Pythagorean theorem.
Read off the outer ratio, and give it the sign that trig function has in $\theta$ 's quadrant.

Worked example: A negative input

Find the exact value of $\tan\left(\arcsin\left(-\dfrac{5}{13}\right)\right)$ .

Solution. Let $\theta = \arcsin\left(-\dfrac{5}{13}\right)$ , so $\sin\theta = -\dfrac{5}{13}$ and $-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$ . The sine is negative, so $\theta$ is in Quadrant IV.

The reference triangle has opposite side $5$ and hypotenuse $13$ , so the adjacent side is $\sqrt{13^2 - 5^2} = \sqrt{144} = 12$ .

In Quadrant IV, $y$ is negative and $x$ is positive. So the opposite side counts as $-5$ and the adjacent side as $12$ :

\tan\theta = \frac{-5}{12} = -\frac{5}{12}.

Worked example: An arccosine in Quadrant II

Find the exact value of $\sin\left(\arccos\left(-\dfrac{1}{3}\right)\right)$ .

Solution. Let $\theta = \arccos\left(-\dfrac{1}{3}\right)$ , so $\cos\theta = -\dfrac{1}{3}$ with $0 \le \theta \le \pi$ . The cosine is negative, so $\theta$ is in Quadrant II, where sine is positive.

The reference triangle has adjacent side $1$ and hypotenuse $3$ , so the opposite side is $\sqrt{9 - 1} = \sqrt{8} = 2\sqrt{2}$ . Therefore

\sin\left(\arccos\left(-\frac{1}{3}\right)\right) = \frac{2\sqrt{2}}{3}.

Tip

A sine of an arccosine, or a cosine of an arcsine, is never negative. Arccosine lands in Quadrants I and II, where sine is positive or zero. Arcsine lands in Quadrants I and IV, where cosine is positive or zero. That's a quick way to catch sign errors.

Writing compositions algebraically

The triangle method also works when the input is a variable. To rewrite $\cos(\arctan x)$ for $x > 0$ , let $\theta = \arctan x$ , so $\tan\theta = \dfrac{x}{1}$ . Draw a triangle with opposite side $x$ and adjacent side $1$ . The hypotenuse is $\sqrt{x^2 + 1}$ , so

\cos(\arctan x) = \frac{1}{\sqrt{x^2 + 1}}.

(This formula also holds for $x \le 0$ , since arctangent lands in Quadrant I or IV, where cosine is positive.) Expressions like this appear often in calculus, where they turn trig expressions back into algebra.