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For each measure, draw an angle with its vertex at the origin of the coordinate plane. Use the positive x-axis as one ray of the angle.
1. 90° 2. 45° 3. 30°
4. 150° 5. 135° 6. 120°
Solutions
1. 2. 3.
4. 5. 6.
For each measure, draw an angle with its vertex at the origin of the coordinate plane. Use the positive x-axis as one ray of the angle.
1. 90° 2. 45° 3. 30°
4. 150° 5. 135° 6. 120°
1-1
1
-1
The Unit Circle-Radius is always one unit-Center is always at the origin-Points on the unit circle relate to the periodic function
30
Let’s pick a point on the unit circle. The positive angle always goes counter-clockwise from the x-axis.
The x-coordinate of this has a value of the cosine of the angle. The y-coordinate has a value of the sine of the angle.
In order to determine the sine and cosine we need a right triangle.
1-1
1
-1
The angle can also be negative. If the angle is negative, it is drawn clockwise from the x axis.
- 45
Find the measure of the angle.
Since 90 + 60 = 150, the measure of the angle is 150°.
The angle measures 60° more than a right angle of 90°.
Sketch each angle in standard position.
a. 48° b. 310° c. –170°
Draw each angle of the unit circle.
a.45o
b.-280 o
c.-560 o
Definition: A circle centered at the origin with a radius of exactly one unit.
|-------1-------|(0 , 0) (1,0)(-1,0)
(0, 1)
(0, -1)
180°
90°
270°
0°
360° 2π
π/2
π
3π/2
0
1-1
1
-1
Let’s look at an example
30
The x-coordinate of this has a value of the cosine of the angle. The y-coordinate has a value of the sine of the angle.
In order to determine the sine and cosine we need a right triangle.
1-1
1
-1
30
Create a right triangle, using the following rules:1.The radius of the circle is the hypotenuse.2.One leg of the triangle MUST be on the x axis. 3.The second leg is parallel to the y axis.
30
601
Remember the ratios of a 30-60-90 triangle-
2
1-1
1
-1
30
30
601
2
X- coordinate
Y- coordinate
P
Find the cosine and sine of 135°.
Use a 45°-45°-90° triangle to find sin 135°.
From the figure, the x-coordinate of point A
is – , so cos 135° = – , or about –0.71. 22
22
opposite leg = adjacent leg
0.71 Simplify.
= Substitute. 22
The coordinates of the point at which the terminal side of a 135° angle intersects are about (–0.71, 0.71), so cos 13 –0.71 and sin 135° 0.71.
Find the exact values of cos (–150°) and sin (–150°).
Step 1: Sketch an angle of –150° in standard position. Sketch a unit circle.
x-coordinate = cos (–150°)y-coordinate = sin (–150°)
Step 2: Sketch a right triangle. Place the hypotenuse on the terminal side of the angle. Place one leg on the x-axis. (The other leg will be parallel to the y-axis.)
(continued)
The triangle contains angles of 30°, 60°, and 90°.
Step 3: Find the length of each side of the triangle.
hypotenuse = 1 The hypotenuse is a radius of the unit circle.
shorter leg = The shorter leg is half the hypotenuse.12
12
32longer leg = 3 = The longer leg is 3 times the shorter leg.
32
12
Since the point lies in Quadrant III, both coordinates are negative. The longer leg lies along the x-axis, so
cos (–150°) = – , and sin (–150°) = – .
Draw each Unit Circle. Then find the cosine and sine of each angle.
a.45o
b.120o
Remember that the unit circle is overlayed on a coordinate plane (that’s how we got the original coordinates for the 90°, 180°, etc.)
Use the side lengths we labeled on the QI triangle to determine coordinates.
45°135°
315°225°
( , )( , )
( , ) ( , )2
2−
2
2
2
2−2
2
2
2−2
2−2
2
2
2
2
2
2
2
π/4
3π/4
5π/4 7π/4
Holding the triangle with the single fold down and double fold to the left, label each side on the triangle.
Unfold the triangle (so it looks like a butterfly) and glue it to the white circle with the triangle you just labeled in quadrant I, on top of the blue butterfly.
Use the side lengths we labeled on the QI triangle to determine coordinates.
60°120°
300°240°
( , )( , )
( , ) ( , )2
3−
2
3
2
1−2
3
2
1−2
3−2
1
2
1
2
1
2
3
π/32π/3
4π/3 5π/3
Holding the triangle with the single fold down and double fold to the left, label each side on the triangle.
Unfold the triangle (so it looks like a butterfly) and glue it to the white circle with the triangle you just labeled in quadrant I, on top of the green butterfly.
We know that the quadrant one angle formed by the triangle is 30°.
That means each other triangle is showing a reference angle of 30°. What about in radians?
Label the remaining three angles.
30°150°
330°210°
π/6
7π/6
5π/6
11π/6
Use the side lengths we labeled on the QI triangle to determine coordinates.
30°150°
330°210°
( , )( , )
( , ) ( , )2
1−
2
1
2
3−2
1
2
3−2
1−2
3
2
3
2
3
2
1
π/6
7π/6
5π/6
11π/6
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