The unit circle provides a convenient way to represent the sine and cosine of special angles in each quadrant.
We first recall the following:
The -coordinate of any point on the unit circle equals the cosine of the corresponding central angle, and
the -coordinate equals the sine of the corresponding central angle.
Let's use the unit circle to list the special values of sine and cosine, starting with the first quadrant. First, we mark off the special angles in this quadrant.
Now, let's write down each point's - and -coordinates, corresponding to the cosine and sine of each point's central angle. In addition, we'll highlight the -coordinate at each point.
Before we move on, observe that if we rotate counterclockwise from the positive -axis, we see that the -coordinates at the special angles are given by the following sequence:
as shown below.
There is a similar pattern for the -coordinates (corresponding to the cosine of the central angle). The only difference is that now we go in reverse, from to as shown below
Take a moment to ensure you can reproduce this diagram on your own without looking. We'll add to it shortly.
So, we know how the unit circle can be used to list the cosine and sine of the special angles in the first quadrant. But what about the other quadrants?
The values in the remaining quadrants can be obtained from the coordinates in the first quadrant. Let's see how:
To obtain the points in the second quadrant, let's first mark off the special angles in this quadrant. These angles have the same reference angles as the angles in the first quadrant:
We mark these angles below and include the quadrantal angle
Now, we reflect the coordinates in the first quadrant in the -axis. This changes the sign of the -coordinate at each point and leaves the -coordinate unchanged.To obtain the points in the third quadrant, we first find the special angles in this quadrant. Again, these angles have the same reference angles as the special angles in the first quadrant:
Then, we take the points in the second quadrant and reflect them in the -axis. This changes the sign of the -coordinate at each point and leaves the -coordinate unchanged.
Finally, to obtain the points in the fourth quadrant, we first compute the special angles in this quadrant:
Then, we take the points in the first quadrant and reflect them in the -axis. This changes the sign of the -coordinate at each point and leaves the -coordinate unchanged.
Therefore, our complete unit circle is as follows.
Take a moment to ensure you can create this diagram yourself without looking. We'll be referring to it often!
What is the value of
Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.
From the unit circle, we see that the angle corresponds to the point
Since the cosine of the angle corresponds to the -coordinate, we have
$\cos 120^\circ =$
|
a
|
$\dfrac 1 2$ |
|
b
|
$-\dfrac{1}{2}$ |
|
c
|
$\dfrac{\sqrt 3}{2}$ |
|
d
|
$-1$ |
|
e
|
$-\dfrac{\sqrt 2}{2}$ |
$\sin 330^\circ =$
|
a
|
$-\dfrac{1}{\sqrt2}$ |
|
b
|
$\dfrac{\sqrt 2}{2}$ |
|
c
|
$\dfrac{\sqrt{3}}{2}$ |
|
d
|
$-\dfrac{\sqrt 3}{2}$ |
|
e
|
$-\dfrac{1}{2}$ |
What is the value of
Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.
From the unit circle, we see that the angle corresponds to the point
Since the sine of the angle corresponds to the -coordinate, we have
$\sin \left(\dfrac {4\pi}{3}\right) =$
|
a
|
$\dfrac {3\sqrt{2}} {4}$ |
|
b
|
$-\dfrac {\sqrt{3}}{2}$ |
|
c
|
$\dfrac {\sqrt{3}}{2}$ |
|
d
|
$\dfrac {2\sqrt{3}}{3}$ |
|
e
|
$-\dfrac {2\sqrt{3}}{3}$ |
$\cos\left( \dfrac {3\pi} {4}\right) =$
|
a
|
$-\dfrac {\sqrt{3}} {2}$ |
|
b
|
$\dfrac {\sqrt{2}} {2}$ |
|
c
|
$-\dfrac {\sqrt{2}} {2}$ |
|
d
|
$\dfrac {\sqrt{3}} {2}$ |
|
e
|
$\dfrac {-1} {2}$ |
Find the value of
Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.
From the unit circle, we see that the angle corresponds to the point
Since the sine of the angle corresponds to the -coordinate, we have
Therefore, using the fact that we have
$\csc 225^\circ =$
|
a
|
$\dfrac{ \sqrt{2}}{2}$ |
|
b
|
$\dfrac{ \sqrt{2}}3$ |
|
c
|
$\dfrac{ \sqrt{3}}{2}$ |
|
d
|
$ -\sqrt{2}$ |
|
e
|
$ \sqrt{3}$ |
$\sec \left(\dfrac {2\pi}{3}\right) =$
|
a
|
$-\dfrac{1}{2}$ |
|
b
|
$-2$ |
|
c
|
$\dfrac{2\sqrt{3}}{3}$ |
|
d
|
$-1$ |
|
e
|
$\dfrac{2}{\sqrt{3}}$ |
Find the value of
Let's remind ourselves of the special trigonometric ratios, as given by the unit circle.
From the unit circle, we see that the angle corresponds to the point
Since the cosine of the angle corresponds to the -coordinate, we have
Since the sine of the angle corresponds to the -coordinate, we have
Therefore, using the fact that we have
$\cot \left(\dfrac {\pi} {3} \right)=$
|
a
|
$-\sqrt{3}$ |
|
b
|
$1$ |
|
c
|
$\sqrt{3}$ |
|
d
|
$\dfrac{1}{\sqrt{3}}$ |
|
e
|
$-\dfrac{1}{\sqrt{3}}$ |
$\tan \left(\dfrac {5\pi} {6} \right)=$
|
a
|
$\dfrac {\sqrt{3}} {3}$ |
|
b
|
$-1$ |
|
c
|
$-\dfrac {\sqrt{3}} {2}$ |
|
d
|
$-\dfrac {\sqrt{2}} {2}$ |
|
e
|
$-\dfrac {\sqrt{3}} 3$ |
