Compositions Involving Inverse Trig Functions, precalculus lesson

A ramp climbs 3 meters for every 4 meters of horizontal run, so its angle is $\theta = \arctan \frac{3}{4}$ . You never need that angle in degrees to know its sine. The right triangle with legs 3 and 4 has hypotenuse 5, so $\sin \theta = \frac{3}{5}$ exactly. Expressions that put a trigonometric function and an inverse trigonometric function together come up often in calculus, and the main difficulty is the restricted range each inverse function returns.

Key terms

Principal ranges

Definition Each inverse trigonometric function returns exactly one angle, taken from a fixed set of angles called its principal range. In radians,

Function	Domain	Principal range
$\arcsin x$	$-1 \le x \le 1$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\arccos x$	$-1 \le x \le 1$	$[0, \pi]$
$\arctan x$	all real $x$	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\operatorname{arcsec} x$	$\lvert x \rvert \ge 1$	$\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$
$\operatorname{arccsc} x$	$\lvert x \rvert \ge 1$	$\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$

So $\theta = \arcsin x$ means both $\sin \theta = x$ and $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ , and the same pattern holds for the others. The ranges for arcsec and arccsc are the convention used in this course, and with them, for every $x$ with $\lvert x \rvert \ge 1$ ,

\begin{gathered} \operatorname{arcsec} x = \arccos \frac{1}{x}, \\[0.8em] \operatorname{arccsc} x = \arcsin \frac{1}{x}. \end{gathered}

Some books choose other ranges for these two.

In plain words An equation like $\sin \theta = \frac{1}{2}$ has infinitely many solutions, and the inverse function picks exactly one of them. The principal range is the rule that says which one, and it tells you the quadrant of the answer before you compute anything.

Example The angles with sine $-\frac{1}{2}$ include $-\frac{\pi}{6}$ , $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ . Only $-\frac{\pi}{6}$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , so $\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ . For cosine the answer looks different. The range $[0, \pi]$ contains no negative angles, so $\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ , an angle in the second quadrant.

Trig function after its inverse

Definition A composition of a trigonometric function with an inverse trigonometric function applies one after the other. When the inverse acts first,

\begin{gathered} \sin(\arcsin x) = x \\[0.4em] \text{and} \quad \cos(\arccos x) = x \\[0.4em] \text{for every } x \text{ in } [-1, 1], \end{gathered}

and

\begin{gathered} \tan(\arctan x) = x \\[0.4em] \text{for every real } x. \end{gathered}

For $\lvert x \rvert > 1$ the expressions $\sin(\arcsin x)$ and $\cos(\arccos x)$ are undefined, because $x$ is not in the domain of the inner function.

In plain words The inverse function finds an angle with the given sine, cosine or tangent, and the outer function reads that value back. This order always returns the input, as long as the inner function accepts it.

Example Reading back gives $\cos\left(\arccos\left(-\frac{2}{3}\right)\right) = -\frac{2}{3}$ and $\tan(\arctan 50) = 50$ . On the other hand $\sin(\arcsin 2)$ has no value at all, since no real angle has sine $2$ .

Inverse after the trig function

Definition The equations

\begin{gathered} \arcsin(\sin x) = x, \\[0.4em] \arccos(\cos x) = x, \\[0.4em] \arctan(\tan x) = x \end{gathered}

hold exactly when $x$ lies in the principal range of the inverse function, that is, for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$ , for $0 \le x \le \pi$ and for $-\frac{\pi}{2} < x < \frac{\pi}{2}$ respectively. For any other $x$ at which the expression is defined, its value is the one angle in the principal range that has the same sine, cosine or tangent as $x$ .

In plain words Taking the sine forgets which of many angles you started with. The inverse function can only return the angle in its own range, so you get $x$ back only when $x$ was already there.

Example The angle $\frac{5\pi}{6}$ lies outside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , and its sine is $\frac{1}{2}$ . So

\arcsin\left(\sin \frac{5\pi}{6}\right) = \arcsin \frac{1}{2} = \frac{\pi}{6}.

In the same way $\tan \frac{3\pi}{4} = -1$ , which gives $\arctan\left(\tan \frac{3\pi}{4}\right) = -\frac{\pi}{4}$ .

Exact evaluation

Definition Exact evaluation gives a value as an exact expression, such as a fraction, a radical or a rational multiple of $\pi$ , rather than as a rounded decimal. To evaluate a composition such as $\cos(\arcsin x)$ exactly, name the inner angle $\theta = \arcsin x$ and write down the two facts it carries,

\sin \theta = x \quad \text{and} \quad -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}.

Then compute the outer function of $\theta$ from those facts, by recognizing a special angle, by a Pythagorean identity or by a reference triangle.

In plain words Give the inner angle a name. The inverse function tells you one trigonometric value of that angle and the interval it lives in, and that is enough to find any other value.

Example Let $\theta = \arccos\left(-\frac{\sqrt{2}}{2}\right)$ . Then $\cos \theta = -\frac{\sqrt{2}}{2}$ with $0 \le \theta \le \pi$ , so $\theta = \frac{3\pi}{4}$ and

\begin{aligned} \sin\left(\arccos\left(-\frac{\sqrt{2}}{2}\right)\right) &= \sin \frac{3\pi}{4} \\[0.8em] &= \frac{\sqrt{2}}{2}. \end{aligned}

No special angle is needed for $\cos\left(\arcsin \frac{1}{3}\right)$ . With $\theta = \arcsin \frac{1}{3}$ the identity gives $\cos^2 \theta = 1 - \frac{1}{9} = \frac{8}{9}$ , and cosine is never negative on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , so

\cos\left(\arcsin \frac{1}{3}\right) = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}.

Reference triangle

Definition When the inverse function returns an angle $\theta$ with $0 < \theta < \frac{\pi}{2}$ , which happens for $\arcsin x$ and $\arccos x$ with $0 < x < 1$ and for $\arctan x$ with $x > 0$ , the angle $\theta$ is an acute angle of a right triangle. A reference triangle for $\theta$ is a right triangle whose sides match the known ratio, for example opposite side $a$ and hypotenuse $c$ when $\sin \theta = \frac{a}{c}$ . The Pythagorean theorem gives the third side, and every other trigonometric value of $\theta$ is a ratio of two sides.

In plain words Turn the input into a triangle. A sine of $\frac{2}{7}$ means opposite side 2 and hypotenuse 7, and once the third side is known you can read off anything.

Example Let $\theta = \arcsin \frac{2}{7}$ . The input is positive, so $\theta$ is acute. Draw a right triangle with opposite side $2$ and hypotenuse $7$ . The adjacent side is $\sqrt{49 - 4} = 3\sqrt{5}$ , and so

\tan\left(\arcsin \frac{2}{7}\right) = \frac{2}{3\sqrt{5}} = \frac{2\sqrt{5}}{15}.

Signs from the quadrant

Definition The principal range places the angle $\theta$ in a quadrant decided by the sign of the input.

Angle	Input between $0$ and $1$	Input between $-1$ and $0$
$\theta = \arcsin x$	Quadrant I	Quadrant IV, $-\frac{\pi}{2} < \theta < 0$
$\theta = \arccos x$	Quadrant I	Quadrant II, $\frac{\pi}{2} < \theta < \pi$

Arctan follows the arcsin pattern for inputs of any size. Every positive input gives an angle in Quadrant I and every negative input an angle in Quadrant IV. The inputs $0$ and $\pm 1$ of arcsin and arccos, and the input $0$ of arctan, give angles on an axis, such as $\arcsin(-1) = -\frac{\pi}{2}$ and $\arccos(-1) = \pi$ . For a negative input, build the reference triangle from the absolute value of the input, find the size of the answer from it, and then give the answer the sign its function has in that quadrant.

In plain words The triangle gives the size of the answer and the quadrant gives the sign. Arcsin and arctan of a negative number point into Quadrant IV, while arccos of a negative number points into Quadrant II. The one exception is the input $-1$ for arcsin and arccos, which lands on an axis.

Example Let $\theta = \arccos\left(-\frac{3}{5}\right)$ , an angle in Quadrant II. The reference triangle has adjacent side 3, hypotenuse 5 and opposite side 4. In Quadrant II sine is positive and tangent is negative, so

\begin{gathered} \sin \theta = \frac{4}{5}, \\[0.8em] \tan\left(\arccos\left(-\frac{3}{5}\right)\right) = -\frac{4}{3}. \end{gathered}

Compare $\varphi = \arcsin\left(-\frac{3}{5}\right)$ , which lies in Quadrant IV. There cosine is positive, so $\cos \varphi = \frac{4}{5}$ .

Algebraic expressions in x

Definition A composition of a trigonometric function with an inverse trigonometric function of a variable can be written without trigonometry. For every $x$ in $[-1, 1]$ ,

\begin{gathered} \cos(\arcsin x) = \sqrt{1 - x^2}, \\[0.4em] \sin(\arccos x) = \sqrt{1 - x^2}, \end{gathered}

for every $x$ in $(-1, 1)$ ,

\tan(\arcsin x) = \frac{x}{\sqrt{1 - x^2}},

and for every real $x$ ,

\begin{gathered} \sin(\arctan x) = \frac{x}{\sqrt{1 + x^2}}, \\[0.8em] \cos(\arctan x) = \frac{1}{\sqrt{1 + x^2}}. \end{gathered}

In plain words These are the reference triangle answers with a letter in place of a number. The principal ranges make the same formulas correct for negative $x$ too, so no separate cases are needed.

Example Take $\theta = \arctan x$ . For $x > 0$ draw a triangle with opposite side $x$ and adjacent side $1$ , so the hypotenuse is $\sqrt{1 + x^2}$ and $\sin \theta = \frac{x}{\sqrt{1 + x^2}}$ . For $x < 0$ the angle is in Quadrant IV, where sine is negative, and the formula is negative too. At $x = -1$ it gives

\sin(\arctan(-1)) = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2},

which matches $\sin\left(-\frac{\pi}{4}\right)$ .

Absolute values

Definition For every real number $u$ ,

\sqrt{u^2} = \lvert u \rvert.

So the Pythagorean identities give, wherever both sides are defined,

\begin{gathered} \sqrt{1 - \sin^2 \theta} = \lvert \cos \theta \rvert, \\[0.4em] \sqrt{1 + \tan^2 \theta} = \lvert \sec \theta \rvert, \\[0.4em] \sqrt{\sec^2 \theta - 1} = \lvert \tan \theta \rvert. \end{gathered}

The absolute value bars may be dropped only when the range of $\theta$ guarantees that the quantity inside them is not negative.

In plain words A square root never comes out negative, so taking the root of a square loses the sign. You get the sign back from the principal range, not from the algebra.

Example Let $\theta = \operatorname{arcsec} x$ with $\lvert x \rvert \ge 1$ . Then $\sec \theta = x$ and $\tan^2 \theta = x^2 - 1$ , so $\lvert \tan \theta \rvert = \sqrt{x^2 - 1}$ . Tangent is not negative on $\left[0, \frac{\pi}{2}\right)$ and not positive on $\left(\frac{\pi}{2}, \pi\right]$ , so

\begin{aligned} &\tan(\operatorname{arcsec} x) \\[0.4em] &\quad = \begin{cases} \sqrt{x^2 - 1} & \text{if } x \ge 1, \\ -\sqrt{x^2 - 1} & \text{if } x \le -1. \end{cases} \end{aligned}

At $x = -2$ the angle is $\operatorname{arcsec}(-2) = \frac{2\pi}{3}$ , and indeed $\tan \frac{2\pi}{3} = -\sqrt{3}$ .

Branch-dependent simplification

Definition A simplification is branch-dependent when the simplified form changes from one interval of the variable to another, because a principal range or an absolute value is involved. For example, for every integer $k$ ,

\begin{gathered} \arctan(\tan x) = x - k\pi \\[0.4em] \text{for } -\frac{\pi}{2} + k\pi < x < \frac{\pi}{2} + k\pi, \end{gathered}

and

\begin{aligned} &\arcsin(\sin x) \\[0.4em] &\quad = \begin{cases} x & \text{if } -\frac{\pi}{2} \le x \le \frac{\pi}{2}, \\ \pi - x & \text{if } \frac{\pi}{2} \le x \le \frac{3\pi}{2}, \end{cases} \end{aligned}

with the same pattern repeating every $2\pi$ .

In plain words One formula does not fit every input. Split the line into pieces and, on each piece, find the expression that has the right sine, cosine or tangent and also lands in the principal range.

Example For $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$ the number $\pi - x$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\sin(\pi - x) = \sin x$ , which is why $\arcsin(\sin x) = \pi - x$ there. Test it at $x = \frac{4\pi}{3}$ . The sine is $-\frac{\sqrt{3}}{2}$ , so

\begin{aligned} \arcsin\left(\sin \frac{4\pi}{3}\right) &= \arcsin\left(-\frac{\sqrt{3}}{2}\right) \\[0.8em] &= -\frac{\pi}{3} = \pi - \frac{4\pi}{3}. \end{aligned}

Check yourself

Find the exact value of $\arccos\left(\cos \frac{7\pi}{6}\right)$ .

Show the answer

The cosine of $\frac{7\pi}{6}$ is $-\frac{\sqrt{3}}{2}$ . The angle in $[0, \pi]$ with this cosine is $\frac{5\pi}{6}$ , so

\arccos\left(\cos \frac{7\pi}{6}\right) = \frac{5\pi}{6}.

Find $\sec\left(\arctan\left(-\frac{5}{12}\right)\right)$ and $\sin\left(\arctan\left(-\frac{5}{12}\right)\right)$ .

Show the answer

The angle $\theta = \arctan\left(-\frac{5}{12}\right)$ lies in Quadrant IV. Its reference triangle has opposite side 5, adjacent side 12 and hypotenuse 13. In Quadrant IV cosine and secant are positive and sine is negative, so

\sec \theta = \frac{13}{12}, \qquad \sin \theta = -\frac{5}{13}.

Write $\tan(\arccos x)$ as an algebraic expression in $x$ , and say for which $x$ it holds.

Show the answer

Let $\theta = \arccos x$ , so $\cos \theta = x$ and $0 \le \theta \le \pi$ . Sine is not negative on $[0, \pi]$ , so $\sin \theta = \sqrt{1 - x^2}$ , and

\begin{gathered} \tan(\arccos x) = \frac{\sqrt{1 - x^2}}{x}, \\[0.8em] -1 \le x \le 1, \; x \neq 0. \end{gathered}

At $x = 0$ the angle is $\frac{\pi}{2}$ , where tangent is undefined. For $-1 < x < 0$ the formula is negative, as it should be in Quadrant II.

Compositions Involving Inverse Trig Functions

Key terms

Principal ranges

Trig function after its inverse

Inverse after the trig function

Exact evaluation

Reference triangle

Signs from the quadrant

Algebraic expressions in x

Absolute values

Branch-dependent simplification

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus