Evaluate the expressions,
answer in degrees and radians.
Notice all the expressions involve
inverse cotangent or arc cotangent.
Which for review has a domain or input of all real numbers
which are all the possible cotangent function values
and their range or output is from zero
to pi radians not including the end points.
Which means in standard position, the output or range
again will be from zero to pi radians not
including the end points.
So, these open circles indicate
the endpoints are not included.
Each of these expressions are equal
to some angle in this interval.
On the unit circle of cotangent theta is equal
to y divided by x, and therefore we will have to determine
this quotient using the points on the unit circle
in order to evaluate the given expressions.
Because we have to find this quotient,
evaluating expressions involving inverse cotangent,
as well as inverse tangent tend to be more challenging,
than evaluating the other four inverse trig functions.
The first expression is inverse cotangent of negative one.
This is equal to an angle in this interval
that has a cotangent function value of negative one.
We should be able to recognize that the quotient
x divided by y is equal to negative one
when x and y are opposites.
So, let's try to find the point
on the unit circle in this interval
where the x and y coordinates are opposites.
Notice that at this point,
x equals a negative square root divided by two,
and y is equal to positive square root two divided by two.
And, therefore x divided by y
is going to equal negative one.
And, therefore a cotangent is equal to negative one
when the terminal side of the angle
intersects the unit circle at this point.
We know the initial side is along the positive x axis.
And, therefore the angle in this interval
that has a cotangent function value
of negative one is this angle here.
135 degrees or 3/4 pi radians.
Next we have inverse cotangent of square root three.
Before we evaluate this, notice all the fractional values
on the unit circle have a denominator of two.
When finding the quotient of two fractions
that have a common denominator,
we can simply find the quotient of the numerators.
Let's quickly show why this works.
If we have a over b divided by c over b,
where we have a common denomenator of b,
we can multiply both fractions by b.
When we do this,
notice how b divided by b simplifies the one here,
as well as here, leaving us
with a simplified fraction of just a over c,
or a divided by c.
And, notice a divided by c is just a quotient
of the numerators from the original fractions.
Again, this only works though when we're dividing fractions
with the same denominator,
which we do on the unit circle.
So, to evaluate inverse cotangent
of the square root of three,
we are looking for the point on the unit circle
in this interval where x divided by y is equal
to the square root of three.
Let's focus on the quotient of the numerators.
Notice that this point here,
if we divide the numerators x divided by y,
the square root three divided by one,
which is equal to square root three.
And, therefore this is the point on the circle
where x divided by y is equal to the square root of three.
And, therefore the angle we are looking for must have a
terminal side intersecting the unit circle at this point.
We know the initial side is here,
and therefore the angle is 30 degrees,
or in radians pi over six radians, or 1/6th pi radians.
But, let's go ahead and show x divided by y anyway.
X divided by y is equal to square root three
over two divided by one half.
Again, to simplify,
we can multiply the top and bottom by two,
simplifying two divided by two simplifies
the one here as well as here,
which does give us the square root three over one,
which is square root three.
This verifies the cotangent function value
is square root three at the angle of 30 degrees,
or 1/6th pi radians.
Next we have inverse cotangent of zero.
Now we're looking for the point on the unit circle
in this interval where x divided by y is equal to zero.
X divided by y is equal to zero at this point here.
Where x is zero and y is one.
Notice x divided by y is equal to zero
divided by one, which is zero.
And, therefore this is the terminal side
of the angle we are looking for.
This is the initial side,
and therefore the expression simplifies to 90 degrees,
or in radians pi over 2 radians, or 1/2 pi radians.
The last expression is the inverse cotangent
of negative square root of three divided by three.
In order to recognize the point where x divided
by y is equal to this expression,
it will be helpful to rationalize the numerator.
Let's go ahead and do that.
We have negative square root three divided by three,
and now to rationalize the numerator,
we multiply the numerator and denominator
by the square root of three.
Notice how this will give us negative three
divided by three square root of three,
then three over three simplifies to one leaving us
with negative one divided by the square root of three.
We can rewrite inverse cotangent
of negative square root three divided by three
as inverse cotangent of negative one divided
by the square root of three.
Again, it's often helpful to change the form
of the given function value,
or in this case, rationalize the numerator
in order to recognize the function value on the unit circle.
So, now we go back to the unit circle
and find a point in the interval where x divided by y
is equal to negative one divided
by the square root of three.
And, again we can focus on the quotient of the numerators.
Notice at this point here x divided by y is negative one
divided by the square root of three,
which is the cotangent function,
and value we are looking for,
and therefore this is the point
on the unit circle we're looking for,
which means this is the terminal side of the angle.
Of course, this is the initial side,
and therefore the angle is 120 degrees,
or in radians 2/3rds pi radians.
I hope you found this helpful.