Introduction

Using a calculator, we can evaluate sine for the angle of 20^\circ, and get

\sin(20^{\circ}) \approx 0.342.

But what if we are given the value 0.342, and we want to find the angle whose sine is this value? To do this, we can use the inverse trigonometric functions.

In the case of the sine, its inverse is arcsine (written as \arcsin or \sin^{-1} ). If applied to the sine, it gives the value of the angle: \arcsin (0.342) \approx 20^{\circ}

The inverses of the other trigonometric functions work in the same way.

  • The inverse of sine is \arcsin.

  • The inverse of cosine is \arccos.

  • The inverse of tangent is \arctan.

Try it for yourself. But make sure that your calculator is in "degrees" mode!

Note: In this topic, we only deal with acute angles. Things get a little more complicated for non-acute angles, but you'll learn about that later!

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Example: Finding the Value of an Inverse Trigonometric Ratio

Find the value of \arcsin (0.2) in degrees to the nearest degree.

EXPLANATION

Using a calculator, we find

\begin{align} \arcsin (0.2) &=(11.536...)^{\circ} \\ &\approx12^{\circ} \end{align} rounded to the nearest integer.

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Practice: Finding the Value of an Inverse Trigonometric Ratio

Find the value of $\arctan\left(\dfrac{1}{8}\right)$ in degrees to the nearest degree.

a
 $9^{\circ}$
b
 $8^{\circ}$
c
 $6^{\circ}$
d
 $7^{\circ}$
e
 $10^{\circ}$
Practice: Finding the Value of an Inverse Trigonometric Ratio

Using a calculator, find the value of $\arccos\left(\dfrac{1}{4}\right)$ in degrees to the nearest degree.

a
 $76^{\circ}$
b
 $36^{\circ}$
c
 $70^{\circ}$
d
 $44^{\circ}$
e
 $81^{\circ}$
Applying an Inverse Trigonometric Function to Both Sides of an Equation

If we have a trigonometric equation like

\cos \theta = \dfrac{1}{2}

and we want to solve for \theta, we can apply the corresponding inverse trigonometric function to both sides of the equation to "cancel" out the trigonometric function.

Here, the trigonometric function is \cos, so the inverse trigonometric function is \arccos, and we have

\begin{align} \arccos \left( \cos \theta \right) &= \arccos \left( \dfrac{1}{2} \right) \\ \theta &= 60^\circ . \end{align}

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Example: Calculating Unknown Angles Using Inverse Trigonometric Functions

If \cos\theta=0.64 and \tan\alpha=1.25 , where \theta and \alpha are acute angles measured in degrees, find \theta+\alpha in degrees to the nearest integer.

EXPLANATION

To find \theta and \alpha we use \arccos and \arctan respectively. We get

\eqalign{ \cos\theta & = 0.64\\ \arccos(\cos\theta) & = \arccos(0.64) \\ \theta&=\arccos(0.64) }

and

\eqalign{ \tan\alpha & = 1.25 \\ \arctan(\tan\alpha) & =\arctan(1.25) \\ \alpha&=\arctan(1.25). }

Then, using a calculator, we find \eqalign{ \theta + \alpha & =\arccos(0.64)+\arctan(1.25)\\ &=(50.208...)^{\circ} + (51.340...)^{\circ} \\ & = (101.548...)^{\circ} \\ & \approx 102^{\circ} }

rounded to the nearest integer.

Watch out! If we round \theta and \alpha to the nearest integer, we get \theta \approx 50^\circ and \alpha \approx 51^\circ. After we sum up the angles we get an incorrect result: \begin{align} \theta + \alpha & \approx 50^\circ + 51^\circ =101^\circ \end{align}

The correct answer is 102^\circ, not 101^\circ. This is a rounding error! To avoid this kind of error we have to round after we take the sum, not before.

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Practice: Calculating Unknown Angles Using Inverse Trigonometric Functions

If $\cos\theta=0.2$ and $\sin\alpha=0.4$, where $\theta$ and $\alpha$ are acute angles measured in degrees, then $\theta+\alpha$ in degrees to one decimal place is

a
 $133.7^{\circ}$
b
 $110.1^{\circ}$
c
 $140.3^{\circ}$
d
 $100.5^{\circ}$
e
 $102.0^{\circ}$
Practice: Calculating Unknown Angles Using Inverse Trigonometric Functions

If $\tan\theta=0.25$ and $\sin\alpha=0.41$, where $\theta$ and $\alpha$ are acute angles measured in degrees, then $\theta+\alpha$ in degrees to one decimal place is

a
 $15.5^{\circ}$
b
 $41.3^{\circ}$
c
 $33.2^{\circ}$
d
 $52.4^{\circ}$
e
 $38.2^{\circ}$
Finding the Measure of an Angle in a Right Triangle Using Trigonometry

The inverse trigonometric functions are particularly useful when we need to find the measure of an angle in a right triangle.

To demonstrate, let's find the measure of the angle \theta indicated in the triangle below.

Since we are given the length of the hypotenuse and opposite sides to \theta, we can use sine to find the angle measurement. We have

\eqalign{ \sin (\theta) = \dfrac{\textrm{opposite}}{\textrm{hypotenuse}} &= \dfrac{5}{7}. }

We now calculate \theta using the inverse sine:

\eqalign{ \sin (\theta) &= \dfrac{5}{7}\\[5pt] \arcsin( \sin (\theta)) &= \arcsin\left(\dfrac{5}{7} \right)\\[5pt] \theta &= \arcsin\left(\dfrac{5}{7} \right)\\[5pt] &=(45.584\,6...)^\circ \\[5pt] &\approx 46^\circ } rounded to the nearest integer.

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Example: Finding the Measure of an Angle Given the Lengths of Two Sides of a Right Triangle

In a right triangle \triangle ABC, we have AC = 9, BC=13, and m\angle A = \alpha. Given that \overline{AB} is the hypotenuse of the triangle, find the value of \alpha (rounded to one decimal place).

EXPLANATION

Let's sketch the triangle that's described.

Since we are given the lengths of the adjacent and opposite sides, we can use tangent to find the angle measurement. We have

\tan \alpha= \dfrac{\textrm{opposite}}{\textrm{adjacent}} = \dfrac{13}{9}.

We now calculate \alpha using the inverse tangent:

\eqalign{ \tan \alpha &= \dfrac{13}{9}\\[5pt] \arctan( \tan \alpha) &= \arctan\left(\dfrac{13}{9} \right)\\[5pt] \alpha &= \arctan\left(\dfrac{13}{9} \right)\\[5pt] &=(55.304\,8...)^\circ\\[5pt] &\approx 55.3^\circ } rounded to one decimal place.

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Practice: Finding the Measure of an Angle Given the Lengths of Two Sides of a Right Triangle

For a right triangle $\triangle ABC,$ $AB = 7, BC = 7\sqrt 3$ and $m\angle A = \theta.$ Given that $\overline{AC}$ is the hypotenuse of the triangle, the value of $\theta$ is

a
 $30^\circ$
b
 $45^\circ$
c
 $75^\circ$
d
 $60^\circ$
e
 $15^\circ$
Practice: Finding the Measure of an Angle Given the Lengths of Two Sides of a Right Triangle

For a right triangle $\triangle ABC,$ $AB = 7, AC = 19$ and $m\angle A = \theta.$ Given that $\overline{AC}$ is the hypotenuse of the triangle, the value of $\theta$ (rounded to one decimal place) is

a
 $71.6^\circ$
b
 $68.4^\circ$
c
 $70.2^\circ$
d
 $52.7^\circ$
e
 $62.1^\circ$
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