We want to solve the equation
tangent theta equals negative one
and the equation tangent theta equals square root of three
divided by three on the interval from zero to two pi
without the use of a calculator,
so we'll be using reference triangles and the unit circle.
So looking at our first equation,
tangent theta equals negative one,
let's first determine which quadrants
tangent theta would be negative.
Remember, tangent theta is equal to y divided by x,
so when x and y have opposite signs,
this quotient would be negative,
and therefore, tangent would be negative,
which means tangent theta would be negative
in the second quadrant,
where the x-coordinate is negative
and the y-coordinate is positive,
and tangent theta would also be negative
in the fourth quadrant,
where the x-coordinate is positive
and the y-coordinate is negative.
Now looking at our equation,
tangent theta equals negative one,
let's ignore the negative sign for a moment
and think about an angle where tangent theta
would be equal to positive one.
Remember, in a right triangle,
tangent theta equals the ratio of the opposite side
to the adjacent side.
And therefore, we should recognize
that tangent 45 degrees, or pi over four radians,
is equal to one divided by one, or positive one.
Again, tangent pi over four radians
is equal to the ratio of the opposite side
to the adjacent side when divided by one,
which equals positive one.
Which means if our angle had a reference angle
of 45 degrees in the second and fourth quadrants,
the tangent function value wouldn't be positive one,
it would be negative one.
So again, any angle that terminates along this ray here
and forms a 45 degree reference angle with the x-axis
would have a tangent function value of negative one,
as well as an angle that terminates
here in the fourth quadrant
that has a 45 degree reference angle.
I know I'm using degrees here,
but once we find the angles in degrees,
we'll convert to radians.
If we were to sketch our reference triangles,
they would look like this
and like this.
And again, because we have 45-45-90 reference triangles,
we normally label the legs one
and the hypotenuse square root two,
but because we're in the second quadrant
where x is negative, this'd be negative one.
In the fourth quadrant,
the y-coordinate is negative,
so this would be negative one.
So the angles we're looking for
over the interval from zero to two pi
would be this angle in the second quadrant
and this angle in the fourth quadrant.
So tangent theta
equals negative one
over this interval when theta equals
90 plus 45, or 135 degrees,
and when theta equals,
this would be 270 plus 45,
315 degrees.
Now let's go ahead and convert these to radians
by multiplying by pi over 180,
and there are four 45s in 180,
three 45s in 135,
so one angle would be three pi over four radians,
and again, there are four 45s in 180,
and there are seven 45s in 315.
So our second solution over the given interval
would be seven pi over four radians.
So these are our two solutions
that we found using our reference triangles.
Let's also show how we can find these on the unit circle.
Because we're looking for
a tangent function value of negative one,
we'd be looking for the points on the unit circle
where the x- and y-coordinates are opposites.
Notice that would occur here.
y divided by x would be negative one
at three pi over four radians,
and y divided by x would also be negative one here
at seven pi over four radians.
Now let's take a look at our second example.
We have tangent theta equals
square root three divided by three.
Let's begin by determining where tangent theta
would be positive.
Well, all trig functions are positive in the first quadrant,
where both x and y are positive,
but y divided by x would also be positive
in the third quadrant,
where both coordinates are negative.
Now, we want to find an angle that has
a tangent function value of square root three
divided by three without using the calculator.
To recognize that angle,
we're probably going to want to change the form
of square root three divided by three.
Let's rationalize this numerator.
So if we have square root three divided by three,
and rationalize the numerator
by multiplying by the square root of three
over square root three,
this would give us three divided
by three square root of three,
which simplifies to one divided by square root three.
Now we want to find an angle
that has a tangent function value
of one divided by square root three,
which should remind us of a 30-60-90 reference triangle,
which we see here.
Notice that tangent
of 30 degrees, or pi over six radians,
again is equal to the ratio of the opposite side
to the adjacent side,
would be one divided by square root three.
Which means if we sketch a 30 degree reference angle
in the first and third quadrants,
we can find our two solutions on the given interval.
So we'll sketch a 30 degree reference angle
in the first quadrant here,
which of course is pi over six radians,
as well as here in the third quadrant.
Let's go ahead and sketch the reference triangles as well.
So we label these short legs one,
the hypotenuse two, the longer legs square root three.
Everything's positive in the first quadrant,
but in the third quadrant,
because x and y are negative,
this would be negative square root three
and negative one.
So the angles we're looking for over this interval
from zero to two pi
again would be here, 30 degrees,
and here, which would be 180 plus 30,
or 210 degrees.
So tangent theta
equals square root three divided by three,
or if we want, one divided by square root three.
We would have theta equals 30 degrees,
which we already know is pi over six radians.
Theta can also be 210 degrees,
which if we convert to radians,
there are six 30s in 180
and seven 30s in 210,
we have seven pi over six radians.
So these are the two solutions on the given interval.
Let's see if we can find these on the unit circle.
We would determine the points where y divided by x
is equal to one divided by square root three.
We notice here, at pi over six radians,
y divided by x would be 1/2
divided by square root three divided by two,
and if we multiply this by two over two,
we would get one over square root three.
And also at 210 degrees, or seven pi over 6 radians,
y divided by x would also be
one divided by square root three.
I hope you found these two examples helpful.