7.3: Inverse Trigonometric Functions

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In previous sections, we have evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that for a one-to-one function, if $f(a)=b$, then an inverse function would satisfy $f^{-1} (b)=a$.

You probably are already recognizing an issue – that the sine, cosine, and tangent functions are not one-to-one functions. To define an inverse of these functions, we will need to restrict the domain of these functions to yield a new function that is one-to-one. We choose a domain for each function that includes the angle zero.

$A portion of the sine graph. This portion starts out at negative pi over 2 comma negative 1, increases concave up to the origin, then increases concave down to pi over 1 comma 1.$ $A portion of the cosine graph. This portion starts out at 0 comma 1, decreases concave down to pi over 2 comma 0, then decreases concave up to pi comma negative 1.$ $A portion of the tangent graph. This portion is one increasing segment, which approaches negative infinity as x approaches negative pi over 2 from the right, and increases concave down up to the origin, then increases concave up, approaching infinity as x approaches pi over 2 from the left.$

On these restricted domains, we can define the inverse sine, inverse cosine, and inverse tangent functions.

INVERSE SINE, COSINE, AND TANGENT FUNCTIONS and their inverses

For angles in the interval $\left[-\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right]$, if $\sin \left(\theta \right)=a$, then $\sin ^{-1} \left(a\right)=\theta$

$\sin ^{-1} \left(x\right)$ has domain [-1, 1] and range $\left[-\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right]$

The $\sin ^{-1} \left(x\right)$ is sometimes called the arcsine function, and notated $\arcsin \left(a\right)$.

For angles in the interval $\left[0,\pi \right]$, if $\cos \left(\theta \right)=a$, then $\cos ^{-1} \left(a\right)=\theta$

$\cos ^{-1} \left(x\right)$ has domain [-1, 1] and range $\left[0,\pi \right]$

The $\cos ^{-1} \left(x\right)$ is sometimes called the arccosine function, and notated $\arccos \left(a\right)$.

For angles in the interval $\left(-\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right)$, if $\tan \left(\theta \right)=a$, then $\tan ^{-1} \left(a\right)=\theta$

$\tan ^{-1} \left(x\right)$ has domain of all real numbers and range $\left(-\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right)$

The $\tan ^{-1} \left(x\right)$ is sometimes called the arctangent function, and notated $\arctan \left(a\right)$.

The graphs of the inverse functions are shown here:

$The graph of the sine inverse function. It starts at negative 1 comma negative pi over 2, increases concave down to the origin, then increases concave up to 1 comma pi over 2.$ $The graph of the cosine inverse function. It starts at negative 1 comma pi, decreases concave up to 0 comma pi over 2, then decreases concave down to 1 comma 0.$ $A graph of the tangent inverse function. It begins just above a horizontal asymptote at negative pi over 2, increasing concave up to the origin, then increase concave down up to and leveling off at another horizontal asymptote at pi over 2.$

Notice that the output of each of these inverse functions is an $angle$.

Example $\PageIndex{1}$

Evaluate

$\sin ^{-1} \left(\dfrac{1}{2} \right)$
$\sin ^{-1} \left(-\dfrac{\sqrt{2} }{2} \right)$
$\cos ^{-1} \left(-\dfrac{\sqrt{3} }{2} \right)$
$\tan ^{-1} \left(1\right)$

Solution

a) Evaluating $\sin ^{-1} \left(\dfrac{1}{2} \right)$ is the same as asking what angle would have a sine value of $\dfrac{1}{2}$. In other words, what angle $\theta$ would satisfy $\sin \left(\theta \right)=\dfrac{1}{2}$?

There are multiple angles that would satisfy this relationship, such as $\dfrac{\pi }{6}$ and $\dfrac{5\pi }{6}$, but we know we need the angle in the range of $\sin ^{-1} \left(x\right)$, the interval $\left[-\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right]$, so the answer will be \[\sin ^{-1} \left(\dfrac{1}{2} \right)=\dfrac{\pi }{6}\nonumber\]

Remember that the inverse is a function so for each input, we will get exactly one output.

b) Evaluating $\sin ^{-1} \left(-\dfrac{\sqrt{2} }{2} \right)$, we know that $\dfrac{5\pi }{4}$ and $\dfrac{7\pi }{4}$ both have a sine value of $-\dfrac{\sqrt{2} }{2}$, but neither is in the interval $\left[-\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right]$. For that, we need the negative angle coterminal with $\dfrac{7\pi }{4}$. \[\sin ^{-1} \left(-\dfrac{\sqrt{2} }{2} \right)=-\dfrac{\pi }{4}\nonumber\]

c) Evaluating $\cos ^{-1} \left(-\dfrac{\sqrt{3} }{2} \right)$, we are looking for an angle in the interval $\left[0,\pi \right]$ with a cosine value of $-\dfrac{\sqrt{3} }{2}$. The angle that satisfies this is \[\cos ^{-1} \left(-\dfrac{\sqrt{3} }{2} \right)=\dfrac{5\pi }{6}\nonumber\]

d) Evaluating $\tan ^{-1} \left(1\right)$, we are looking for an angle in the interval $\left(-\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right)$ with a tangent value of 1. The correct angle is \[\tan ^{-1} \left(1\right)=\dfrac{\pi }{4}\nonumber\]

Exercise $\PageIndex{1}$

Evaluate

$\sin ^{-1} \left(-1\right)$
$\tan ^{-1} \left(-1\right)$
$\cos ^{-1} \left(-1\right)$
$\cos ^{-1} \left(\dfrac{1}{2} \right)$

Answer

a) $-\dfrac{\pi }{2}$

b) $-\dfrac{\pi }{4}$

c) $\pi$

d) $\dfrac{\pi }{3}$

Example $\PageIndex{2}$

Evaluate $\sin ^{-1} \left(0.97\right)$ using your calculator.

Solution

Since the output of the inverse function is an angle, your calculator will give you a degree value if in degree mode, and a radian value if in radian mode.

In radian mode, \[\sin ^{-1} (0.97) \approx 1.3252\nonumber\]

In degree mode, \[\sin ^{-1} \left(0.97\right)\approx 75.93{}^\circ\nonumber\]

Exercise

Evaluate $\cos ^{-1} \left(-0.4\right)$ using your calculator.

Answer: \[1.9823\text{ or }113.578\mathrm{{}^\circ}\nonumber\]

In Section 5.5, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trig functions, we can solve for the angles of a right triangle given two sides.

Example $\PageIndex{3}$

Solve the triangle for the angle $\theta$.

$A right triangle with hypotenuse labeled 12 and one side labeled 9, and the angle between them labeled theta.$

Solution

Since we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.

\[\cos \left(\theta \right)=\dfrac{9}{12}\nonumber\]Using the definition of the inverse,

\[\theta =\cos ^{-1} \left(\dfrac{9}{12} \right)\nonumber\]Evaluating

\[\theta \approx 0.7227\text{, or about }41.4096\mathrm{{}^\circ}\nonumber\]

There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can find exact values for the resulting expressions

Example $\PageIndex{4}$

Evaluate $\sin ^{-1} \left(\cos \left(\dfrac{13\pi }{6} \right)\right)$.

Solution

a) Here, we can directly evaluate the inside of the composition.

\[\cos \left(\dfrac{13\pi }{6} \right)=\dfrac{\sqrt{3} }{2}\nonumber\]

Now, we can evaluate the inverse function as we did earlier.

\[\sin ^{-1} \left(\dfrac{\sqrt{3} }{2} \right)=\dfrac{\pi }{3}\nonumber\]

Exercise $\PageIndex{3}$

Evaluate $\cos ^{-1} \left(\sin \left(-\dfrac{11\pi }{4} \right)\right)$.

Answer: \[\sin \left(-\dfrac{11\pi }{4} \right)=-\dfrac{\sqrt{2} }{2} . \cos ^{-1} \left(-\dfrac{\sqrt{2} }{2} \right)=\dfrac{3\pi }{4} \nonumber\]

Example $\PageIndex{5}$

Find an exact value for $\sin \left(\cos ^{-1} \left(\dfrac{4}{5} \right)\right)$.

Solution

Beginning with the inside, we can say there is some angle so $\theta =\cos ^{-1} \left(\dfrac{4}{5} \right)$, which means $\cos \left(\theta \right)=\dfrac{4}{5}$, and we are looking for $\sin \left(\theta \right)$. We can use the Pythagorean identity to do this.

\[\sin ^{2} \left(\theta \right)+\cos ^{2} \left(\theta \right)=1\nonumber\]Using our known value for cosine
\[\sin ^{2} \left(\theta \right)+\left(\dfrac{4}{5} \right)^{2} =1\nonumber\]Solving for sine
\[\sin ^{2} \left(\theta \right)=1-\dfrac{16}{25}\nonumber\]
\[\sin \left(\theta \right)=\pm \sqrt{\dfrac{9}{25} } =\pm \dfrac{3}{5}\nonumber\]

Since we know that the inverse cosine always gives an angle on the interval $\left[0,\pi \right]$, we know that the sine of that angle must be positive, so \[\sin \left(\cos ^{-1} \left(\dfrac{4}{5} \right)\right)=\sin (\theta )=\dfrac{3}{5}\nonumber\]

Example $\PageIndex{6}$

Find an exact value for $\sin \left(\tan ^{-1} \left(\dfrac{7}{4} \right)\right)$.

Solution

While we could use a similar te $A right triangle with legs 4 and 7. The side length 4 meets the hypotenuse at an angle labeled theta.$ chnique as in the last example, we will demonstrate a different technique here. From the inside, we know there is an angle so $\tan \left(\theta \right)=\dfrac{7}{4}$. We can envision this as the opposite and adjacent sides on a right triangle.

Using the Pythagorean Theorem, we can find the hypotenuse of this triangle:

\[4^{2} +7^{2} =hypotenuse ^{2}\nonumber\]

\[hypotenuse=\sqrt{65}\nonumber\]

Now, we can represent the sine of the angle as opposite side divided by hypotenuse.

\[\sin \left(\theta \right)=\dfrac{7}{\sqrt{65} }\nonumber\]

This gives us our desired composition

\[\sin \left(\tan ^{-1} \left(\dfrac{7}{4} \right)\right)=\sin (\theta )=\dfrac{7}{\sqrt{65} } .\nonumber\]

Exercise $\PageIndex{4}$

Evaluate $\cos \left(\sin ^{-1} \left(\dfrac{7}{9} \right)\right)$.

Answer

Let $\theta =\sin ^{-1} \left(\dfrac{7}{9} \right)$ so \[\sin (\theta )=\dfrac{7}{9}\nonumber\]

Using Pythagorean Identity, $\sin ^{2} \theta +\cos ^{2} \theta =1$, so \[\left(\dfrac{7}{9} \right)^{2} +\cos ^{2} \theta =1\nonumber\]

Solving, \[\cos \left(\sin ^{-1} \left(\dfrac{7}{9} \right)\right)=\cos \left(\theta \right)=\dfrac{4\sqrt{2} }{9}\nonumber\]

We can also find compositions involving algebraic expressions

Example $\PageIndex{7}$

Find a simplified expression for $\cos \left(\sin ^{-1} \left(\dfrac{x}{3} \right)\right)$, for $-3\le x\le 3$.

Solution

We know there is an angle $\theta$ so that $\sin \left(\theta \right)=\dfrac{x}{3}$. Using the Pythagorean Theorem,

\[\sin ^{2} \left(\theta \right)+\cos ^{2} \left(\theta \right)=1\nonumber\]Using our known expression for sine
\[\left(\dfrac{x}{3} \right)^{2} +\cos ^{2} \left(\theta \right)=1\nonumber\]Solving for cosine
\[\cos ^{2} \left(\theta \right)=1-\dfrac{x^{2} }{9}\nonumber\]
\[\cos \left(\theta \right)=\pm \sqrt{\dfrac{9-x^{2} }{9} } =\pm \dfrac{\sqrt{9-x^{2} } }{3}\nonumber\]

Since we know that the inverse sine must give an angle on the interval $\left[-\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right]$, we can deduce that the cosine of that angle must be positive. This gives us

\[\cos \left(\sin ^{-1} \left(\dfrac{x}{3} \right)\right)=\dfrac{\sqrt{9-x^{2} } }{3}\nonumber\]

Exercise $\PageIndex{5}$

Find a simplified expression for $\sin \left(\tan ^{-1} \left(4x\right)\right)$, for $-\dfrac{1}{4} \le x\le \dfrac{1}{4}$.

Answer

Let $\theta =\tan ^{-1} \left(4x\right)$, so $\tan (\theta )=4x$. We can represent this on a triangle as $\tan (\theta )=\dfrac{4x}{1}$. $A right triangle with one angle labeled theta, and the leg opposite labeled 4x and the leg adjacent labeled 1.$

The hypotenuse of the triangle would be $\sqrt{\left(4x\right)^{2} +1}$. \[\sin \left(\tan ^{-1} \left(4x\right)\right)=\sin (\theta )=\dfrac{4x}{\sqrt{16x^{2} +1} }\nonumber\]

Important Topics of This Section

Inverse trig functions: arcsine, arccosine and arctangent
Domain restrictions
Evaluating inverses using unit circle values and the calculator
Simplifying numerical expressions involving the inverse trig functions
Simplifying algebraic expressions involving the inverse trig functions

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Important Topics of This Section