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CCSS.Math: ,

Let's say that we're told
that some angle theta, which is going to be
expressed in radians, is between negative 3 pi
over 2 and negative pi. It's greater than
negative 3 pi over 2 and it's less than negative pi. And we're also told that sine
of theta is equal to 1/2. Just from this
information can we figure out what the tangent of
theta is going to be equal to? And I encourage you to pause the
video and try this on your own. In case you're stumped,
I will give you a hint. You should use the Pythagorean
identity, the fact that sine squared theta, plus cosine
squared theta is equal to 1. So let's do it. So we know the Pythagorean
identity, sine squared theta, plus cosine squared
theta is equal to 1. We know what sine
squared theta is. Sine theta is 1/2. So this could be
rewritten as 1/2 squared, plus cosine squared
theta, is equal to 1. Or we could write this as
1/4 plus cosine squared theta is equal to 1. Or we could subtract
1/4 from both sides, and we get cosine squared
theta is equal to-- let's see. You subtract 1/4 from
the left hand side, then this 1/4 goes away. That was the whole point. 1 minus 1/4 is 3/4. So what could
cosine of theta be? Well, when I square
it, I get positive 3/4. So it could be the positive or
negative square root of 3/4. So cosine of theta
could be equal to the positive or negative
square root of 3 over 4, which is the same thing as
the positive or negative square root of 3, over the
square root of 4, which is 2. So it's the positive or negative
square root of 3 over 2. But how do we know which
one of these it actually is? Well, that's where this
information becomes useful. Let's draw our unit circle. If you're saying,
well, why am I even worried about cosine of theta? Well, if you know sine of
theta you know cosine of theta. Tangent of theta is just sine
of theta over cosine theta. So then you will know
the tangent of theta. But let's look at
the unit circle to figure out which value
of cosine we should use. So let me draw it,
the unit circle. That's my y-axis. That is my x-axis. And I will draw the
unit circle in pink. So that's my best attempt
at drawing a circle. Please forgive me for its
lack of perfect roundness. And it says theta is greater
than negative 3 pi over 2. So where is negative
3 pi over 2? So let's see. This is negative pi over 2. So this is one
side of the angle. Let me do this in a color. So this one side of
the angle is going to be along the positive x-axis. And we want to figure out
where the other side is. So this right over here
that's negative pi over 2. This is negative pi. So it's between negative pi,
which is right over here. So let me make that clear. Negative pi is right over here. It's between negative pi
and negative 3 pi over 2. Negative 3 pi over 2
is right over here. So our angle theta is going
to put us someplace over here. And the whole reason I did
this-- so this whole arc right here-- you
could think of this as the measure of angle
theta right over there. And the whole
reason I did that is to think about whether
the cosine of theta is going to be
positive or negative. We clearly see it's in
the second quadrant. The cosine of theta
is the x-coordinate of this point where our angle
intersects the unit circle. So this point right
over here-- actually let me do it in
that orange color again-- this right over here,
that is the cosine of theta. Now is that a positive
or negative value? Well it's clearly
a negative value. So for the sake of this
example, our cosine theta is not a positive 1. It is a negative 1
So we could write the cosine theta is equal to
the negative square root of 3 over 2. So we figured out cosine
theta, but we still have to figure out
tangent of theta. And we just have
to remind ourselves that the tangent
of theta is going to be equal to the sine of
theta over the cosine of theta. Well they told us the
sine of theta is 1/2. So it's going to be 1/2
over cosine of theta, which is negative square
root of 3 over 2. And what does that equal? Well that's the same thing
as 1/2 times the reciprocal of this. So times negative 2 over
the square root of 3. These twos will cancel out and
we are left with negative 1 over the square root of 3. Now some people don't like a
radical in the denominator, like this. They don't like an
irrational denominator. So we could rationalize
the denominator here by multiplying by square root
of 3 over square root of 3. And so this will be equal
to negative square root of 3 over 3 is the tangent of
this angle right over here. And that actually makes
sense, because the tangent of the angle is the
slope of this line. And we see that it is
indeed a negative slope.