Lesson Video: Solving Equations Using Inverse Trigonometric Functions Mathematics

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

12:46

Video Transcript

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

Before we talk about the inverse trig functions, let’s remind ourselves what we know about trig functions. To do that, let’s consider this figure. We have a right triangle with a vertical length of 𝑦 and a horizontal length of 𝑥. Notice that the distance from the origin to 𝑥, 𝑦 is one. This means that the triangle is inscribed in a circle of radius one. A circle with a center at the origin and a radius of one is known as the unit circle. So, what we see here is a right triangle inscribed in the unit circle. And we will talk about the sine, the cosine, and tangent in relationship to this right triangle in the unit circle. The 𝑥-value represents the adjacent side length. The 𝑦-value represents the opposite side length. And then, we have a hypotenuse of one, the distance from the origin to the point 𝑥, 𝑦.

Given this triangle with an angle of 𝜃, the sin of 𝜃 is equal to the opposite over the hypotenuse. The cos of 𝜃 is equal to the adjacent over the hypotenuse, and the tan of 𝜃 is equal to the opposite over the adjacent. From this information, we get a really important principle. For trig functions, the input value or the domain will be an angle measure and the output will be a ratio of side lengths. We input some 𝜃 and the output will be a ratio. Using this information and what we know about inverse functions, we can say that the output of the trig functions, the range, must become the domain of its inverse, and the input of the trig functions, its domain, will become the range of the inverse trig functions.

The inverse sine, you input a ratio, opposite over hypotenuse, and the output is an angle measure. Cosine inverse, we input an adjacent over hypotenuse side ratio, and we get an angle measure. The same thing is true for tangent. What we’re saying is that for our inverse trig function, the values that we input into the function will be ratios of side lengths and the output will be some angle measure. The output can be in degrees or in radians. But before we go any further, we need to make some distinctions.

Because the three trig functions are not one-to-one functions, they fail the horizontal line test, when we’re working with them, we have to restrict their domain. And that means, in order to operate with the inverses of the trig functions, we’ll also need to be operating under restricted domains. In this case, we’re only going to be operating from zero degrees to 90 degrees or from zero radians to 𝜋 over two radians. We’re going to limit ourselves to working with the inverse functions in the first quadrant. If you’re working with inverses in a calculator, they will automatically be giving you the first quadrant answers. So, let’s consider how we use these inverse trig functions to solve equations.

Given that 𝐴 is an acute angle and sin 𝐴 is equal to 0.8193, determine the measure of angle 𝐴 to the nearest tenth of a degree.

We know that sin of some unknown angle 𝐴 is equal to 0.8193. If we think about this trig function, the 𝜃-value is an angle measure, and it’s equal to some ratio of side lengths. If we know the ratio of side lengths and we want to know the angle measure, we need a function that can undo the sine function, and that would be the sin inverse of the sin 𝜃. We’ll need to take the inverse sin of sin 𝐴. So, we’ll do that to both sides of our equation. The sin inverse of sin 𝐴 is just equal to 𝐴, and the sin inverse of 0.8193 can be found with a calculator. Make sure that your calculator is set to degrees, since we’re interested in this angle measure in degrees.

When we do that, we get 55.0147 continuing degrees. We’ll round this to the nearest tenth, and we’ll see that the measure of angle 𝐴 is 55.0 degrees. We’ve been given that 𝐴 is an acute angle, and so we’re only interested in the smallest value that 𝐴 can be, which here is 55 degrees.

Before we go on, we should know the different notations for inverse trig functions. We’ve already seen sine with the superscript of negative one, and we read that the inverse sine. But this is also sometimes written as the arcsine, and these two notations mean exactly the same thing. The arcsine returns the angle whose sine is a given number. Similarly, the inverse cosine can be written as arccosine and the inverse tangent as arctan. Let’s look at another example where we need to use an inverse trig function to solve an equation.

Find the value of 𝑋 given that the tan of 𝑋 over four is equal to the square root of three, where 𝑋 over four is an acute angle.

We know that we’ll only be dealing with angles in the first quadrant. We also know that the tangent of 𝜃 is equal to the opposite over the adjacent. And that means for some angle 𝜃, the tangent of 𝜃 is equal to this ratio. And for us, that angle is 𝑋 over four, and the ratio is the square root of three. When we see this tan of some angle equals the square root of three, we might recognize that this is a special angle. We know that the tan of 60 degrees equals the square root of three. And if we know that the tan of 60 degrees equals the square root of three, then we can set 𝑋 over four equal to 60 degrees. Then, we multiply both sides of the equation by four and find that 𝑋 must be equal to 240 degrees.

You might all of a sudden be wondering, “Wait! I thought that this was an acute angle.” But we have to be careful here because 𝑋 over four is an acute angle; that doesn’t mean 𝑋 by itself has to be less than 90 degrees. 𝑋 over four is 60 degrees, which is an acute angle, and 𝑋 is 240 degrees.

But now you might be wondering, what if you didn’t remember that tan of 60 degrees equals the square root of three? In this case, you could use the inverse trig function. If you take the inverse tan of tan of 𝑋 over four, you just get 𝑋 over four. And if you take the inverse tan of the square root of three and plug that into a calculator, it will give you 60 degrees. Now you want to make sure that your calculator is operating in degrees and not in radians when you enter the tan inverse of the square root of three. Both methods show that 𝑋 must be equal to 240 degrees in order for 𝑋 over four to equal 60 degrees.

Now, we’re ready to consider another example.

Find the measure of angle 𝑋 in degrees given that two times cos of 𝑋 equals the tan of 60 degrees, where 𝑋 is an acute angle.

Since 𝑋 is an acute angle, we’re only interested in solutions in quadrant one. So, we can look at two times cos of 𝑋 is equal to tan of 60 degrees. And the first thing that we can do is solve for tan of 60 degrees, which is equal to the square root of three. Because this is an angle that we usually memorize, you maybe already knew that the tan of 60 degrees was the square root of three. If not, you could plug that into your calculator. Now, we have two times cos of 𝑋 equals the square root of three. Our goal is to solve for 𝑋. We’re trying to get 𝑋 by itself. So, we divide through by two and we see that the cos of 𝑋 has to equal the square root of three over two.

And one of two things can happen here. You might remember that the cos of 30 degrees is the square root of three over two. Or you might recognize that we can solve for 𝑋 by finding the inverse cosine of both sides of this equation. The inverse cos of the cos of 𝑋 equals 𝑥 and the inverse cos of the square root of three over two can be plugged into a calculator, which will return 30 degrees. If you knew that the cos of 30 degrees was the square root of three over two, you would recognize that 𝑋 has to be equal to 30 degrees. But both methods prove this.

Now, we’re ready to look at one final example.

Find the measure of angle 𝐸 given that the tan of 𝐸 equals 18.5845 and angle 𝐸 is an acute angle. Give your answer to the nearest second.

Since we know that angle 𝐸 is an acute angle, we know that we’ll only be dealing with values in the first quadrant. We know that the tan of 𝐸 is equal to 18.5845. We know that the tangent of some angle equals a ratio of side lengths. And we also know that if we’re given a ratio of side lengths for some tangent, using the inverse tangent, we can find the angle measure. And that means we’ll want to take the inverse tangent of both sides of this equation. The inverse tan of tan of 𝐸 equals 𝐸. And before we find the inverse tan of 18.5845, we need to think about “are we dealing with radians or degrees?” We need to round this value to the nearest second. A second is a fraction of minutes, and a minute is a fraction of a degree. That means we’ll need to operate this inverse tangent in degrees.

Making sure our calculator is set to degrees, we get that 𝐸 equals 86.91998286 continuing degrees. This means 𝐸 has 86 whole degrees and another partial degree. In order for to us to turn this partial degree into minutes and seconds, we have to remember that one degree equals 60 minutes. And that means to find out how many minutes 0.91998286 continuing degrees would be, we multiply by 60. And when we do that, we get 55.19897132 continuing minutes, 55 whole minutes and a partial minute.

Our final step will be converting this partial minute into seconds. And for that, we have to know that one minute equals 60 seconds. So, we have 86 degrees, 55 minutes and then we’ll have 0.19897132 times 60 minutes, which gives us 11.93827 continuing seconds. To round to the nearest second, we’ll round that 11.9 up to 12. And we’ll say that the measure of angle 𝐸 is equal to 86 degrees, 55 minutes, and 12 seconds.

Now, we’re ready to review our key points. For some value sin 𝜃, cos 𝜃, or tan 𝜃 where 𝜃 is an acute angle, we can use the inverse trigonometric functions to find the missing angle 𝜃. We can represent that like this that the sin inverse of sin 𝜃 will equal 𝜃, the cos inverse of cos of 𝜃 equals 𝜃, and the tan inverse of tan of 𝜃 will equal 𝜃.

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