Evaluate the expressions.
Answer in degrees and radians.
Notice all of the expressions involve inverse tangent
or arc tangent, where for review,
has a domain of all real numbers and a range or output
from negative pi over two to positive pi over two radians,
not including the end points.
Which means in standard position,
the output will be from zero to pi over two radians,
not including pi over two,
or from zero to negative pi over two radians,
not including negative pi over two.
On the unit circle, tangent theta is equal to y divided by x
and therefore we will have to find this quotient
to evaluate the given expressions,
which makes evaluating inverse tangent more challenging
that evaluating inverse cosine and inverse sine.
The first expression is inverse tangent of negative one.
This is equal to the angle and the range,
where in this interval it has a tangent function
value of negative one.
But we should be able to recognize
that y divided by x is equal to negative one
when the x and y coordinates are opposites.
Notice that this point on the unit circle,
x is equal to y, and therefore the quotient is positive one.
But at this point on the unit circle,
the x and y coordinates are opposites,
and therefore y divided by x is equal to negative one,
which means the terminal set of the angle we are looking for
must intersect the unit circle at this point.
And we know the initial side must be along
the positive x-axis, but the angle we are looking for
is not 315 degrees or seven-fourths pi radians,
because that angle is not in the range of inverse tangent.
We will have to rotate clockwise,
and therefore the angle is negative 45 degrees,
or in radians, negative pi over four radians,
or negative 1/4 pi radians.
Next we have inverse tangent of square root three.
Before we evaluate this though, notice all the fractions
on the circle have a denominator of two,
which does make determining y divided by x much easier.
Let's review the shortcut.
If we have the fraction A over B divided by C over B,
where both fractions have a denominator of B,
we can clear the denominators of the complex fraction
by multiplying both fractions by B.
Once we do this, notice how B divided by B
simplifies to one here as well as here,
and when simplifying, we end up with just A over C,
or A divided by C, which notice is just the quotient
of the numerators of the fractions.
So when determining y divided by x on the circle,
we can just focus on the quotient of the numerators.
So going back to our examples, let's now determine
the inverse tangent of square root three.
Again we need to find the point on the unit circle
where y divided by x is equal to square root three.
If we focused on the numerators,
notice how this point here, y divided by x
would be square root three divided by one,
which is equal to square root three.
This is the point of the unit circle
where y divided by x is equal to square root of three,
which means the angle we are looking for
has a terminal side that intersects
the unit circle at this point.
The initial side is along the positive x axis,
and therefore the angle we are looking for
is 60 degrees, or in radians,
pi over three radians, or 1/3 pi radians.
But just to be sure, let's go ahead and find y divided by x.
Y divided by x is equal to square root of three over two,
divided by 1/2.
We can simplify the complex fraction by multiplying
the top and bottom by two.
Two divided by two simplifies to one here as well as here,
showing this does equal square root of three over one,
which is square root of three.
So this does verify we have the correct angle.
Next we have inverse tangent
of negative square root of three divided by three.
Sometimes it is helpful to rationalize the numerator
of the given tangent function value
in order to recognize the point on the unit circle.
So for this example, let's rationalize the numerator.
We have negative square root of three over three.
Let's rationalize the numerator by multiplying
the numerator and denominator by the square root of three.
This gives us negative three
over three square root of three.
Three divided by three simplifies to one,
giving us negative one divided by the square root of three.
So we can write inverse tangent
of negative square root of three divided by three
as inverse tangent of negative one
over square root of three, which in this form
should make it easier to recognize
the point on the unit circle we are looking for.
We need to find the point where y divided by x
is equal to negative one divided by square root three.
We know it's going to be in the fourth quadrant.
Focusing on the quotient of the numerators,
notice how here y divided by x is going to be
negative one over square root three,
and therefore this is the point
we're looking for on the unit circle.
The terminal side of the angle must intersect
the unit circle at this point.
The initial side is along the positive x axis.
We need to rotate clockwise,
the angle is negative 30 degrees,
or in radians, negative pi over six radians,
or negative 1/6 pi radians.
And if we need to, we can verify y divided by x.
That's equal to negative 1/2 divided by
the square root of three over two.
To simplify we can multiply the top and bottom by two.
Simplifying, two divided by two
simplifies to one here and here,
which does give us negative one
over the square root of three.
Now, if we wanted to get this value here
we would have to rationalize the denominator.
Let's also show that.
We'd multiply the top and bottom
by negative square root of three,
which does give us a negative
square root of three over three.
But you can see why it was helpful to rationalize
the numerator here, because it did make it much easier
to recognize the correct point on the unit circle.
For the last expression, we have inverse tangent of zero.
We now look for the point on the circle
where y divided by x is equal to zero,
which is this point here.
Y divided by x is zero divided by one, which is zero.
The terminal side of the angle
must intersect the unit circle at this point,
which is the same side as the initial side,
and therefore the angle is zero degrees, or zero radians.
I hope you found this helpful.