Lesson Explainer: Solving Equations Using Inverse Trigonometric Functions Mathematics

In this explainer, we will learn how to solve equations by using inverse trigonometric functions in the first quadrant.

The first quadrant means we only consider acute angles πœƒ, where 0<πœƒ<90∘∘ or 0<πœƒ<πœ‹2 in radians.

The inverse trigonometric functions have a number of real-world applications in many disciplines including engineering, navigation, physics, and geometry. For example, if you are doing carpentry and you want to make sure that the end of a piece of wood is cut at a particular angle, you can do this by measuring the sides and using an inverse trigonometric function to determine the angle. Or suppose you want to climb a mountain but do not want to climb up an elevation greater than a particular angle; you can measure the horizontal distance to the mountain from your starting position and the distance of your path up the mountain and use inverse trigonometric functions to ensure you do not exceed your maximum elevation angle or to determine which mountain to climb if you want to reach the top.

Before we talk about the inverse trigonometric functions, let’s begin by recalling the trigonometric functions, whose inverses we will examine in this explainer. Consider the right triangle shown.

The trigonometric functions can be expressed in terms of the ratio of the sides of the triangle as sinOHcosAHtanOAπœƒ=,πœƒ=,πœƒ=.

These functions satisfy the following trigonometric identities: sincostansincosοŠ¨οŠ¨πœƒ+πœƒβ‰‘1,πœƒβ‰‘πœƒπœƒ.

The domain (the set of possible inputs) and range (the set of possible outputs) of the trigonometric functions are beyond the scope of this explainer. We will only consider input values in the first quadrant for acute angles, for the trigonometric functions.

The common trigonometric values, in degrees and radians, are given in the following table.

Angle (πœƒ)30ο‚‹πœ‹6∘45ο‚‹πœ‹4∘60ο‚‹πœ‹3∘
sinπœƒ12√22√32
cosπœƒβˆš32√2212
tanπœƒβˆš331√3

We can remember these from the following special right triangles.

We can see the use of the inverse trigonometric functions by considering the following question. Suppose a 14 m ladder is placed against a wall. If the bottom of the ladder is 7 m away from the base of the wall, what angle does the bottom of the ladder make with the ground?

For this problem, the length of the hypotenuse side is 14 m and the length of the side adjacent to the angle πœƒ is 7 m. Since cosine is the ratio between the lengths of the adjacent and hypotenuse in a right triangle, we have cosAHmmπœƒ==714=12.

In order to find the angle πœƒ, which is an acute angle, we can look up the common trigonometric values to see which value of πœƒ satisfies this equation for cosπœƒ. Therefore, the angle made by the bottom of the ladder with the ground is given by πœƒ=60,πœ‹3.∘orradians

In this problem, we found the angle just by inspecting different values of πœƒ for the cosine function. We can then ask the question: what if πœƒ is not a common trigonometric value that we can deduce by working backward? This is where we turn to the inverse trigonometric functions that essentially allow us to make πœƒ the subject of the equation.

The sine, cosine, and tangent trigonometric functions are periodic functions. Their domains have to be restricted to a particular subset, known as the principle branch, in order to have inverse functions. This is not an issue in this explainer, since we restrict the domain to the first quadrant, 0,πœ‹2 in radians or ]0,90[∘∘ in degrees.

Definition: Inverse Trigonometric Functions

The inverse trigonometric functions denoted by sin, cos, and tan are the inverse functions of the trigonometric functions sin, cos, and tan. This means they work in reverse or β€œgo backward” from the usual trigonometric functions. For acute angles, π‘¦βˆˆοŸ0,πœ‹2 in radians or π‘¦βˆˆ]0,90[∘∘ in degrees; they are defined by 𝑦=π‘₯⟺π‘₯=𝑦,𝑦=π‘₯⟺π‘₯=𝑦,𝑦=π‘₯⟺π‘₯=𝑦.sinsincoscostantan

These can also be written as arcsinπ‘₯, arccosπ‘₯, and arctanπ‘₯.

For example, if we have sin(30)=12∘, then this is equivalent to sinοŠ±οŠ§βˆ˜ο€Ό12=30. (The inverse of a function, 𝑓(π‘₯), can also be denoted as 𝑓(π‘₯), so if π‘₯=𝑓(𝑦)=𝑦sin, then 𝑦=𝑓(π‘₯)=π‘₯sin, and vice versa. This will be covered in more detail in another explainer on inverse functions in general.)

From the definition of the inverse function and for acute angles πœƒβˆˆοŸ0,πœ‹2 in radians or ]0,90[∘∘ in degrees, the inverse trigonometric functions satisfy the properties sinsincoscostantan(πœƒ)=πœƒ,(πœƒ)=πœƒ,(πœƒ)=πœƒ.

The input and output values are interchanged for the inverse trigonometric functions from the usual trigonometric functions. For the inverse trigonometric functions, we only consider output values in the first quadrant for acute angles, corresponding to the input values of the trigonometric functions which we mentioned earlier.

We can also express the angle πœƒ in our right triangle in terms of the given sides by using the inverse trigonometric functions: sinOHsinsinsinOHsinOHπœƒ=,(πœƒ)=,πœƒ=, and similarly for the other expressions: πœƒ=ο€Όοˆ,πœƒ=.cosAHtanOA

For our problem with the ladder, we could have used the inverse cosine function to find the angle πœƒ in a similar way: coscoscoscoscosπœƒ=12,(πœƒ)=ο€Ό12,πœƒ=ο€Ό12,πœƒ=60.∘

By using the table for trigonometric values in reverse, we can also create a table for the common inverse trigonometric values as follows.

π‘₯12√22√32
sinπ‘₯30ο‚‹πœ‹6∘45ο‚‹πœ‹4∘60ο‚‹πœ‹3∘
cosπ‘₯60ο‚‹πœ‹3∘45ο‚‹πœ‹4∘30ο‚‹πœ‹6∘
π‘₯√331√3
tanπ‘₯30ο‚‹πœ‹6∘45ο‚‹πœ‹4∘60ο‚‹πœ‹3∘

These types of problems can be solved for any acute angle and not just those given by the common values. Suppose we want to solve the equation 3πœƒβˆ’2.946=0,sin to determine the acute angle πœƒ to the nearest degree. We begin by rearranging the equation to isolate sinπœƒ: 3πœƒβˆ’2.946=03πœƒ=2.946πœƒ=2.9463πœƒ=0.982.sinsinsinsin

Now, we apply the inverse sine function, sin, to make πœƒ subject: sinsinsin∘(πœƒ)=(0.982)πœƒ=79.1125….

We obtained the value of sin(0.982) by using the arcsin function on a calculator, set to degrees mode. Therefore, to the nearest degree, we have πœƒ=79.∘

Now, suppose for an acute angle 𝑋 measured in radians we want to solve the equation 2√3π‘‹βˆ’2𝑋=0.sincos

We want to find the value of 𝑋, in radians, such that this equation is satisfied. We begin by rearranging the equation and using a trigonometric identity to rewrite the expression in terms of tan𝑋: 2√3π‘‹βˆ’2𝑋=02√3𝑋=2𝑋𝑋=1√3𝑋𝑋𝑋=1√3𝑋=√33.sincossincossincossincostan

Finally, we apply the inverse tangent function, arctan, to make 𝑋 the subject and find the appropriate value from the table of common angles or using the arctan function on a calculator, set to radians mode: tantantan(𝑋)=ο€Ώβˆš33,𝑋=πœ‹6.

We will now look at a few examples to practice and deepen our understanding.

For the first example, we will apply the inverse sine function to determine a particular angle, in degrees, to the nearest tenth of a degree.

Example 1: Using Inverse Trigonometric Functions to Solve a Trigonometric Equation

Given that 𝐴 is an acute angle and that sin𝐴=0.8193, determine π‘šβˆ A to the nearest tenth of a degree.

Answer

In this example, we want to determine an angle using the inverse sine function, to the nearest tenth of a degree.

In particular, we will make use of the sine property for acute angles, π‘₯∈]0,90[∘∘: sinsin(π‘₯)=π‘₯.

We begin by taking the inverse sine function sin of both sides of the equation: sinsinsinsin𝐴=0.8193(𝐴)=(0.8193)𝐴=55.0147….∘

We obtained the value of sin(0.8193) by using the arcsin function on a calculator, set to degrees mode.

Therefore, to the nearest tenth of a degree, π‘šβˆ π΄=55.0.∘

In our next example, we will use the inverse cosine function to determine a particular angle in radians.

Example 2: Using Inverse Trigonometric Functions to Solve Trigonometric Equations Involving Special Angles

Given that π‘₯ is an acute angle and 4(π‘₯)=2√3cos, determine the value of π‘₯ in radians.

Answer

In this example, we want to determine an angle using the inverse cosine function, in radians.

In particular, we will make use of the cosine property for acute angles, π‘₯∈0,πœ‹2: coscos((π‘₯))=π‘₯.

We begin by rearranging the equation to make cos(π‘₯) the subject: 4(π‘₯)=2√3(π‘₯)=2√34(π‘₯)=√32.coscoscos

Now, we apply the inverse cosine function cos to both sides of the equation and use the table of common angles or the arccos function on a calculator, set to radians mode, to determine the value of π‘₯: coscoscos((π‘₯))=ο€Ώβˆš32π‘₯=πœ‹6.

In our next example, we will use the inverse cosine function again to determine a particular angle, but this time in degrees by using the knowledge of the tangent of a common angle.

Example 3: Using Inverse Trigonometric Functions to Solve Trigonometric Equations Involving Special Angles

Find the measure of βˆ π‘‹ in degrees given 2𝑋=60costan∘ where 𝑋 is an acute angle.

Answer

In this example, we want to determine an angle using the inverse cosine function.

In particular, we will make use of the cosine property for acute angles, π‘₯∈]0,90[∘∘: coscos(π‘₯)=π‘₯.

First note that tan60=√3∘. We begin by rearranging the equation to isolate cos𝑋: 2𝑋=√3𝑋=√32.coscos

Now, we apply the inverse cosine function cos to both sides of the equation and use the table of common angles or the arccos function on a calculator, set to degrees mode, to determine the value of 𝑋: coscoscos∘(𝑋)=ο€Ώβˆš32𝑋=30.

In our next example, we will use the inverse tangent function to determine a particular angle and give our answer to the nearest second.

Example 4: Using Inverse Trigonometric Functions to Solve Trigonometric Equations

Find π‘šβˆ πΈ given tan𝐸=18.5845 and ∠𝐸 is an acute angle. Give the answer to the nearest second.

Answer

In this example, we want to determine an angle using the inverse tangent function to the nearest second.

In particular, we will make use of the tangent property for acute angles, π‘₯∈]0,90[∘∘: tantan(π‘₯)=π‘₯.

We begin by taking the inverse tangent function of both sides of the equation: tantantantan𝐸=18.5845(𝐸)=(18.5845)𝐸=86.9199….∘

We obtained the value of tan(18.5845) by using the arctan function on a calculator, set to degrees mode.

Now, we convert this angle into the right form by expressing it as 86.91998286…=86+5560+11.9382…60=8655β€²11.9382…′′.∘∘

Therefore, to the nearest second, we have π‘šβˆ πΈ=8655β€²12β€²β€².∘

In our next example, we will use the inverse tangent and inverse cosine functions to solve two equations involving trigonometric functions.

Example 5: Solving Trigonometric Equations Involving Multiple Trigonometric Functions and Special Angles

Find the smallest positive angle that satisfies both 2πœƒβˆ’βˆš2=0cos and tanπœƒβˆ’1=0.

Answer

In this example, we want to solve a pair of equations by using the inverse cosine and tangent functions.

Since we are looking for the smallest positive angle solving these equations, let’s begin by looking for acute solutions. In particular, we will make use of the cosine and tangent property for acute angles, π‘₯∈]0,90[∘∘: coscostantan(π‘₯)=π‘₯,(π‘₯)=π‘₯.

We begin by rearranging the first equation to make cosπœƒ the subject of the equation: 2πœƒβˆ’βˆš2=02πœƒ=√2πœƒ=√22.coscoscos

Now, we can take the inverse cosine function cos of both sides of the equation and use the table of common angles or the arccos function on a calculator, set to degrees mode, to determine the value of πœƒ: coscoscos∘(πœƒ)=ο€Ώβˆš22ο‹πœƒ=45.

We now check that this satisfies the second condition by substituting πœƒ=45∘ into the second equation to check that it is satisfied, or we determine the angle directly from the second equation. For the latter, we begin by rearranging the equation to isolate tanπœƒ: tantanπœƒβˆ’1=0πœƒ=1.

Next, we apply the inverse tangent function tan to both sides of the equation and use the table for the common angles or the arctan function on a calculator, set to degrees mode, to determine the angle πœƒ: tantantan∘(πœƒ)=(1)πœƒ=45.

Thus, the smallest positive angle that satisfies both 2πœƒβˆ’βˆš2=0cos and tanπœƒβˆ’1=0 is πœƒ=45.∘

In our next example, we will use the inverse cosine function to determine a particular angle different from the common angles, in radians.

Example 6: Using Inverse Trigonometric Functions to Solve Trigonometric Equations Involving Special Angles

Given that π‘₯ is an acute angle and 2√2(π‘₯)=1+√3cos, determine the value of π‘₯ in radians.

Answer

In this example, we want to determine the value of an acute angle using the inverse cosine function, in radians.

In particular, we will make use of the cosine property for acute angles, π‘₯∈0,πœ‹2: coscos((π‘₯))=π‘₯.

We begin by rearranging the equation to make cos(π‘₯) the subject: 2√2(π‘₯)=1+√3(π‘₯)=1+√32√2.coscos

Now, we apply the inverse tangent function cos to both sides of the equation and use the arccos function on a calculator, set to radians mode, to determine the angle π‘₯: coscoscos((π‘₯))=ο€Ώ1+√32√2π‘₯=πœ‹12.

In our final example, we will use the inverse tangent function to determine a particular angle.

Example 7: Using Inverse Trigonometric Functions to Solve Trigonometric Equations Involving Special Angles

Find the value of 𝑋 given tan𝑋4=√3 where 𝑋4 is an acute angle.

Answer

In this example, we want to determine an angle using the inverse tangent function.

In particular, we will make use of the tangent property for acute angles, π‘₯∈]0,90[∘∘: tantan(π‘₯)=π‘₯.

We begin by taking the inverse tangent function, tan, of both sides of the equation and use the table for the common angles or the arctan function on a calculator, set to degrees mode, to determine the acute angle 𝑋4: tantantantan𝑋4=√3𝑋4=ο€»βˆš3𝑋4=60.∘

Therefore, the value of 𝑋 is given by 𝑋=240.∘

Key Points

  • We can use the inverse trigonometric functions sin, cos, and tan to solve trigonometric equations.
  • For some values of sinπœƒ, cosπœƒ, or tanπœƒ, where πœƒ is an acute angle, we can use the inverse trigonometric functions to find the missing angle in degrees or radians.
  • The key properties to solve these problems for acute angles πœƒβˆˆ]0,90[∘∘ in degrees or πœƒβˆˆοŸ0,πœ‹2 in radians are sinsincoscostantan(πœƒ)=πœƒ,(πœƒ)=πœƒ,(πœƒ)=πœƒ.

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