Compositions of Inverse Trig Functions

Compositions of inverse trig functions combine a trig function with an inverse trig function, like $\sin(\arcsin x)$ or $\cos(\arctan \frac{3}{4})$ . A function undoes its own inverse when the input is allowed: $\sin(\arcsin x) = x$ . For a mixed pair, draw a right triangle for the inner angle and read the outer ratio from it.

Before this lesson: You should know Inverse Trig Functions, Composition of Functions and the Pythagorean Theorem.

Working From the Inside Out

A composition is evaluated from the inside out, as in composition of functions. In $\cos(\arctan \frac{3}{4})$ , the inner part $\arctan \frac{3}{4}$ is an angle $\theta$ with $\tan \theta = \frac{3}{4}$ . The outer part asks for the cosine of that same angle.

You don’t need to know $\theta$ in degrees. Draw a right triangle where tangent is $\frac{3}{4}$ : opposite 3 and adjacent 4. The Pythagorean theorem gives a hypotenuse of 5, so $\cos \theta = \frac{4}{5}$ .

A right triangle with angle theta at the bottom left. The side opposite theta is 3, the side adjacent to theta is 4 and the hypotenuse is 5. A label says theta equals arctan of 3/4. — If θ = arctan(3/4), then tan θ = 3/4. The triangle gives the other ratios, such as cos θ = 4/5.

Rules for Compositions

Composition	Result	When it works
$\sin(\arcsin x)$	$x$	$-1 \le x \le 1$
$\cos(\arccos x)$	$x$	$-1 \le x \le 1$
$\tan(\arctan x)$	$x$	All real $x$
$\arcsin(\sin \theta)$	$\theta$	Only if $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$
$\arccos(\cos \theta)$	$\theta$	Only if $0 \le \theta \le \pi$
$\arctan(\tan \theta)$	$\theta$	Only if $-\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}$

If the outside and inside match, check the domain or range rule in the table.
If they don’t match, or the angle is outside the range, evaluate the inside first.
For a mixed pair, let $\theta$ be the inner angle and sketch a right triangle from its ratio.
Find the missing side with the Pythagorean theorem.
Read the outer ratio. Give it the sign for the quadrant where $\theta$ lies.

Worked Examples

Example 1: A function and its inverse

Evaluate $\sin(\arcsin 0.4)$ and $\cos(\arccos(-0.7))$ .

Check the domain: 0.4 is between −1 and 1

\sin(\arcsin 0.4) = 0.4

Sine undoes arcsine.

Check the domain: −0.7 is between −1 and 1

\cos(\arccos(-0.7)) = -0.7

Cosine undoes arccosine.

Answer: 0.4 and −0.7

Example 2: When the inverse does not undo

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

\sin \frac{5\pi}{6} = \frac{1}{2}

The angle $\frac{5\pi}{6}$ is outside arcsine’s range, so evaluate the inside first.

\frac{\pi}{6}

Now find $\arcsin \frac{1}{2}$ , the angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ with sine $\frac{1}{2}$ .

Answer: $\frac{\pi}{6}$ , not $\frac{5\pi}{6}$ . The two angles have the same sine, but only $\frac{\pi}{6}$ is in the range of arcsine.

Example 3: Use a right triangle

Evaluate $\cos(\arctan \frac{3}{4})$ .

Let $\theta = \arctan \frac{3}{4}$ , so $\tan \theta = \frac{3}{4}$ and $\theta$ is in Quadrant I

Draw a right triangle with opposite 3 and adjacent 4

c^2 = 3^2 + 4^2 = 9 + 16 = 25

Find the hypotenuse.

c = 5

Take the square root.

\cos \theta = \frac{4}{5}

Read cosine, adjacent over hypotenuse.

Answer: $\cos(\arctan \frac{3}{4}) = \frac{4}{5}$

Example 4: An angle in Quadrant II

Evaluate $\sin(\arccos(-\frac{5}{13}))$ .

Let $\theta = \arccos(-\frac{5}{13})$ . Arccosine of a negative value is in Quadrant II.

In Quadrant II, use $x = -5$ and $r = 13$

25 + y^2 = 169

Find $y$ with $x^2 + y^2 = r^2$ .

Subtract 25 from both sides: $y^2 = 144$ , so $y = 12$ (positive in Quadrant II)

\sin \theta = \frac{y}{r} = \frac{12}{13}

Read sine.

Answer: $\frac{12}{13}$ . This uses the point method from trig functions of any angle.

Example 5: Word problem with a drone

A drone is 100 meters away from you horizontally. Its angle of elevation is $\theta = \arctan \frac{h}{100}$ , where $h$ is its height. When $h = 75$ meters, find $\cos \theta$ and the straight-line distance $d = \frac{100}{\cos \theta}$ to the drone.

\theta = \arctan \frac{75}{100} = \arctan \frac{3}{4}

Substitute.

Draw the triangle: opposite 3, adjacent 4, hypotenuse 5 (scaled from 75, 100 and 125)

\cos \theta = \frac{4}{5}

Read cosine.

d = \frac{100}{\frac{4}{5}} = 100 \cdot \frac{5}{4} = 125

Find the distance.

Answer: $\cos \theta = \frac{4}{5}$ , and the drone is 125 meters away. The Pythagorean theorem agrees: $\sqrt{100^2 + 75^2} = \sqrt{15{,}625} = 125$ .

Common Mistakes

Mistake	Example of the error	Fix
Canceling without checking the range	$\arcsin(\sin \frac{5\pi}{6}) = \frac{5\pi}{6}$	$\frac{5\pi}{6}$ is outside $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . The answer is $\frac{\pi}{6}$ .
Canceling outside the domain	$\sin(\arcsin 2) = 2$	$\arcsin 2$ is undefined, so the composition is undefined.
Wrong sign from the triangle	$\cos(\arcsin(-\frac{8}{17})) = -\frac{15}{17}$	Arcsine of a negative is in Quadrant IV, where cosine is positive: $\frac{15}{17}$ .
Using the wrong sides	For $\arctan \frac{3}{4}$ , using hypotenuse 4	Tangent is opposite over adjacent: opposite 3, adjacent 4.
Multiplying instead of composing	$\cos(\arctan \frac{3}{4}) = \cos \cdot \frac{3}{4}$	Find the angle first, then take its cosine.

Related Methods

Algebraic expressions

The triangle method also works with a variable. For $\cos(\arcsin x)$ , draw opposite $x$ and hypotenuse 1. The adjacent side is $\sqrt{1 - x^2}$ , so $\cos(\arcsin x) = \sqrt{1 - x^2}$ . Calculus uses results like this often.

Check with a calculator

You can check any answer numerically. In radian mode, $\cos(\tan^{-1}(0.75)) = 0.8$ , which matches $\frac{4}{5}$ .

Reciprocal outer functions

If the outer function is secant, cosecant or cotangent, find the matching ratio from the triangle and flip it. For example, $\sec(\arctan 2) = \frac{\sqrt{5}}{1} = \sqrt{5}$ .

Practice Problems

Evaluate $\sin(\arcsin 0.25)$ .
Evaluate $\tan(\arctan 7)$ .
Evaluate $\arccos(\cos \frac{\pi}{3})$ .
Evaluate $\arcsin(\sin \frac{2\pi}{3})$ .
Evaluate $\arccos(\cos(-\frac{\pi}{4}))$ .
Evaluate $\sin(\arctan \frac{5}{12})$ .
Evaluate $\tan(\arccos \frac{3}{5})$ .
Evaluate $\cos(\arcsin(-\frac{8}{17}))$ .
Evaluate $\sec(\arctan 2)$ .
A 10-foot ladder leans against a wall at an angle $\theta = \arccos \frac{6}{10}$ with the ground. Find how high the ladder reaches, $10 \sin \theta$ .

Show answers

0.25 is between −1 and 1, so sine undoes arcsine
Answer: 0.25

Tangent undoes arctangent for every real number
Answer: 7

$\frac{\pi}{3}$ is in $[0, \pi]$ , so arccosine undoes cosine
Answer: $\frac{\pi}{3}$

\sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}

Evaluate the inside.

\arcsin \frac{\sqrt{3}}{2} = \frac{\pi}{3}

Find the arcsine.

Answer: $\frac{\pi}{3}$

\cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2}

Evaluate the inside.

\arccos \frac{\sqrt{2}}{2} = \frac{\pi}{4}

Find the arccosine.

Answer: $\frac{\pi}{4}$

Triangle: opposite 5, adjacent 12

\sqrt{25 + 144} = \sqrt{169} = 13

Find the hypotenuse with the Pythagorean theorem.

\frac{5}{13}

Read sine.

Answer: $\frac{5}{13}$

Triangle: adjacent 3, hypotenuse 5

\sqrt{25 - 9} = \sqrt{16} = 4

Find the opposite side with the Pythagorean theorem.

\frac{4}{3}

Read tangent.

Answer: $\frac{4}{3}$

The angle is in Quadrant IV: $y = -8$ , $r = 17$
Find x: $x^2 = 289 - 64 = 225$ , so $x = 15$ (positive in Quadrant IV)

\frac{15}{17}

Read cosine.

Answer: $\frac{15}{17}$

Triangle: opposite 2, adjacent 1

\sqrt{4 + 1} = \sqrt{5}

Find the hypotenuse with the Pythagorean theorem.

\frac{\sqrt{5}}{1}

Secant is hypotenuse over adjacent.

Answer: $\sqrt{5}$

10.

Triangle: adjacent 6, hypotenuse 10

\sqrt{100 - 36} = \sqrt{64} = 8

Find the opposite side with the Pythagorean theorem.

So $\sin \theta = \frac{8}{10}$

10 \cdot \frac{8}{10} = 8

Find the height: multiply by the sine.

Answer: 8 feet

FAQ

Does sin(arcsin x) always equal x?

Yes, as long as $x$ is between −1 and 1. For any other $x$ , $\arcsin x$ is undefined, so the composition is undefined.

Why doesn’t arcsin(sin x) always equal x?

Arcsine only returns angles from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ . If $x$ is outside that range, arcsine returns a different angle with the same sine.

How do you evaluate cos(arctan x)?

Draw a right triangle with opposite $x$ and adjacent 1. The hypotenuse is $\sqrt{x^2 + 1}$ , so $\cos(\arctan x) = \frac{1}{\sqrt{x^2 + 1}}$ .

How do you know the sign of the answer?

Use the range of the inner inverse function. Arcsine and arctangent of a negative value are in Quadrant IV, and arccosine of a negative value is in Quadrant II. Then apply the quadrant sign rules.

Summary

Evaluate compositions from the inside out.
$\sin(\arcsin x) = x$ and $\cos(\arccos x) = x$ for $-1 \le x \le 1$ ; $\tan(\arctan x) = x$ for all $x$ .
$\arcsin(\sin \theta) = \theta$ only when $\theta$ is in arcsine’s range; otherwise evaluate the inside first.
For mixed pairs, draw a right triangle for the inner angle and read the outer ratio.
Use the inner function’s range to choose the quadrant and sign.

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