### 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.