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Trigonometry – Radian Measure and The Unit Circle
Measurements can be referenced to any number of so-called ‘units’. For example, lengths can
be measured in inches, feet, meters, etc. The same is true for angle measurements. So far, we
have used degrees to measure angles. In this lesson we introduce an alternate unit called
‘radians’. Although it is far more common to use degrees in common speech, radians is much
more common in science and engineering work. We also formally introduce the concept of a
unit circle as it relates to the trigonometric functions.
Radians
Using a rectangular coordinate system, we measure an angle in standard position using the
origin as the vertex and the positive 𝑥-axis as the initial ray. The angle is formed by rotating a
second ray counterclockwise. Choosing a fixed location along the 𝑥-axis, this action sweeps out
an arc, 𝑠, on a circle with radius equal to the starting location on the 𝑥-axis, i.e. 𝑥 = 4.
x
y
θ
4
r
s
The radian is defined in terms of this radius, 𝑟, and the length of the arc, 𝑠, as follows:
Sweeping an angle of 1 radian results in an arc length that is equal to the length of the radius.
Similarly, sweeping an angle of 2 radians results in an arc length that if equal to twice the length
of the radius.
x
y
1 radian
r
s = r
θ
x
y
2 radians
r
s = 2r
θ
With this we can write a simple relationship between the arc length and the radius as follows.
Next, we recall a familiar elementary formula for the circumference of a circle.
The circumference represents the arc length after one full rotation, i.e. 𝑠 = 𝐶 for 𝜃 = 2 𝜋. Since
one full rotation is also known to be equivalent to 360
°
, we can say 360
°
= 2 𝜋 or 180
°
We can use the relationship to convert between degrees and radians as summarized below.
Degrees to Radians Radians to Degrees
Multiply the angle in degree by
𝝅
𝟏𝟖𝟎
°
Example:
°
°
Multiple the angle in radians by
𝟏𝟖𝟎
°
𝝅
Example:
°
°
The unit circle is one of the most important tools for understanding and working with
trigonometric functions. Before moving on to this let’s do some examples with converting
radian and degree angle measures.
Example 1: Convert each degree measure to radians.
a. 45
°
b. − 270
°
c. 249. 8
°
Solution:
a. 45
°
𝜋
180
°
𝜋
4
b. − 270
°
𝜋
180
°
3 𝜋
2
c. 249. 8
°
𝜋
180
°
Example 2 : Convert each radian measure to degree.
a.
9 𝜋
4
b. −
5 𝜋
6
c. 4. 25
Solution:
a.
9 𝜋
4
9 𝜋
4
180
°
𝜋
°
b. −
5 𝜋
6
5 𝜋
6
180
°
𝜋
°
c. 4. 25 = 4. 25 ∙ (
180
°
𝜋
°
This unit circle should be used as a reference when evaluating trigonometric functions.
However, rather than trying to fully commit the figure to memory, I suggest you understand
how to create it from fundamental principles. It contains the same information that was in the
table from the previous lesson, (except for the radian measure), but organized on the unit
circle, which highlights the various symmetries as they apply to the cosine and sine values.
- Symmetric with respect to 𝒙 - axis
Flipping the cosine and sine values about the 𝑥-axis we note that the cosine values, which
refer to the 𝑥-axis, remain the same, and the sine values, which refer to the 𝑦-axis, flip in
sign.
- Symmetric with respect to 𝒚 - axis
Flipping the cosine and sine values about the y - axis we note that the sine values, which
refer to the 𝑦-axis, remain the same, and the cosine values, which refer to the 𝑥-axis, flip in
sign.
We can use these symmetries, along with the ‘trick’ for memorizing the special angles in the
first quadrant we learned two lessons ago, to reproduce the unit circle.
When producing the unit circle its generally more difficult to remember the radian values
compared to the degree values. For this we describe one method that can help.
Imagine dividing the circle into 12 pieces, i.e. 15
°
each. Next, we pull out a factor of 𝜋 from the
radian values giving us a measure that goes from 0 to 2 in steps of 1 ⁄ 12. Knowing this we can
easily reproduce the radian values with simple increments of 1 12
. The figure below shows
the degree and radian values (less the 𝜋). Be sure to verify for yourself, by reducing the
fractions and multiplying by 𝜋, that the values shown here correspond to the values from
above. Note that we sometimes have increments of 30
°
, i.e. 2 / 12 , e.g. between 60
°
and 90
°
45
°
60
°
30
°
0
12
90
°
120
°
135
°
150
°
0
°
180
°
210
°
225
°
240
°
270
°
300
°
315
°
330
°
360
° 24
12
2
12
3
12
4
12
6
12
8
12
9
12
10
12
12
12
14
12
15
12
16
12 18
12
20
12
21
12
22
12
Example 3 : Given the quadrant where 𝜃 exists, find its exact value for each of the following
equations.
a. 𝑄 2 : 𝑠𝑖𝑛
1
2
b. 𝑄 2 : 𝑐𝑜𝑠
1
2
c. 𝑄 3 : 𝑡𝑎𝑛(𝜃) = 3
d. 𝑄 3 : 𝑠𝑒𝑐
2 3
3
e. 𝑄 4 : 𝑡𝑎𝑛
f. 𝑄 4 : 𝑐𝑜𝑠
3
2
Solution: For this example, as well as example 4 and 5, you should start by reproducing the unit
circle without consulting the one given above. When complete verify it with the one from
above and then solve the examples using your unit circle as your reference. To begin drawing
the unit circle it usually helps to create a quick table for the first quadrant using our ‘trick’ from
a previous lesson.
°
°
°
°
°
a. In the first quadrant 𝑠𝑖𝑛
1
2
for 𝜃 = 30
°
. The sine value stays positive above the 𝑥-
axis, i.e. 1
st
and 2
nd
quadrant. In the 2
nd
quadrant 150
°
corresponds to 30
°
. You can
visualize this by imagining flipping the Q1 triangle over the 𝑦-axis. Therefore, in the 2
nd
quadrant 𝑠𝑖𝑛
1
2
for
°
b. In the first quadrant 𝑐𝑜𝑠(𝜃) =
1
2
for 𝜃 = 60
°
. The cosine value is negative on the left side
of the 𝑦-axis, i.e. 2
nd
and 3
rd
quadrant. In the 2
nd
quadrant 120
°
corresponds to 60
°
. You
can visualize this by imagining flipping the Q1 triangle over the y-axis. Therefore, in the 2
nd
quadrant 𝑐𝑜𝑠(𝜃) =
1
2
for
°
Example 4 : Find the exact function values without using a calculator
a. 𝑡𝑎𝑛 (
𝜋
3
) b. 𝑐𝑜𝑠 (
2 𝜋
3
) c. 𝑠𝑖𝑛 (−
5 𝜋
6
d. 𝑡𝑎𝑛 (−
7 𝜋
3
) e. 𝑐𝑠𝑐 (−
11 𝜋
6
) f. 𝑐𝑜𝑡
Solution:
a. Using the table from example 3 we find that
b. In this case we can use our alternate version of the unit circle by removing the 𝜋 term in
the angle and changing the fraction to have a denominator of 12. This gives us an ‘angle’ of
or 8 ∙ 15
°
°
, which is in the 2
nd
quadrant where the cosine value is negative. In
the 2
nd
quadrant 120
°
corresponds to 60
°
. The cosine of 60
°
is 1 2
. Therefore, we have
°
c. We can again start by converting the angle to degrees by removing the 𝜋 term and
changing the fraction to have a denominator of 12, i.e. − 10 12
or − 10 ∙ 15
°
°
Next, we can change to a positive angle using coterminal angles by adding 360
°
°
°
°
°
is in the 3
rd
quadrant where the sine is negative. Furthermore, in the 3
rd
quadrant 210
°
corresponds to 30
°
. The sine of 30
°
is 1 ⁄ 2 , therefore,
°
°
d. We start by changing the angle similar to the previous example, starting with − 28 ⁄ 12. In
degrees we have − 28 ∙ 15
°
°
. Adding 2 ∙ 360
°
we find the coterminal angle.
°
°
°
Which is in the 4
th
quadrant where the tangent is negative. Finally, since 300
°
corresponds to
°
for which the tangent is 3. Therefore,
°
°
e. In this case, − 11 ⁄ 6 = − 22 ⁄ 12 , and − 22 ∙ 15
°
°
. Adding 360
°
we find a
coterminal angle of 30
°
. Since 30
°
is in the 1
st
quadrant we can directly use the simple
table from above to find 𝑠𝑖𝑛
°
. Finally, since 𝑐𝑠𝑐
we have
°
°
f. Each time we rotate 𝜋 = 180
°
we travel halfway around the circle. You will notice that
when we rotate on odd integer number of times, we always end up at 180
°
. Therefore
°
°
°
Example 5 : Find the exact values of 𝜃 over the given interval that satisfy the given conditions.
a.
[
3
2
b.
[
1
2
c.
[
2
1
2
d.
[
2
= 3 e.
[
2
f. [− 2 𝜋,𝜋): 𝑠𝑖𝑛
2
1
2
Solution:
a. From our simple 1
st
quadrant table, we know that 𝑠𝑖𝑛( 60
°
3
2
. We also know that the
sine is negative in quadrants 3 and 4. In the 3
rd
quadrant 60
°
corresponds to 210
°
and in
the 4
th
quadrant it corresponds to 300
°
. Therefore,
3
2
for 𝜃 = [ 210
°
°
] = [
7
6
5
3
𝜋].
b. In this case, 𝑐𝑜𝑠( 60
°
1
2
. Furthermore, the cosine is negative in quadrants 2 and 3. In the
nd
quadrant 60
°
corresponds to 120
°
and in the 3
rd
quadrant it corresponds to 210
°
Therefore, 𝑐𝑜𝑠(𝜃) = −
1
2
for 𝜃 = [ 120
°
°
] = [
2
3
7
6
𝜋].
f. Start again with the square root and find
Similar to problem c. we have
𝜃 = [ 45
°
°
°
°
]
Since the solution range is the same as in problem e., we use the same technique.
°
°
°
°
°
°
°
°
°
°
°
The final solution set is then
[
°
°
°
°
°
°
]
Final Summary for Trigonometry – Radian Measure and The Unit Circle
Degrees and Radians
One complete revolution around a circle is 360
°
or 2 𝜋 radians. Knowing this we can convert between
these two angle measures as shown below.
Degrees to Radians Radians to Degrees
Multiply the angle in degree by
𝝅
𝟏𝟖𝟎
°
Example:
°
°
Multiple the angle in radians by
𝟏𝟖𝟎
°
𝝅
Example:
°
°
The Unit Circle
The unit circle can be used to solve basic trigonometric equations which are related to the
special angles, i.e. 0
°
°
°
°
°
45
°
,
𝜋
4
2
2
,
2
2
1
2
,
3
2
60
°
,
𝜋
3
30
°
,
𝜋
6
3
2
,
1
2
90
°
,
𝜋
2
120
°
,
2 𝜋
3
135
°
,
3 𝜋
4
150
°
,
5 𝜋
6
( 0 , 1
)
−
1
2
,
3
2
−
2
2
,
2
2
−
3
2
,
1
2
0
°
, 0
360
°
, 2 𝜋
( 1 , 0
180 )
°
(− 1 , 0 ) , 𝜋
210
°
,
7 𝜋
6
225
°
,
5 𝜋
4
240
°
,
4 𝜋
3
270
°
,
3 𝜋
2
( 0 , − 1
)
300
°
,
5 𝜋
3
315
°
,
7 𝜋
4
330
°
,
11 𝜋
6
3
2
, −
1
2
2
2
, −
2
2
1
2
, −
3
2
−
3
2
, −
1
2
−
2
2
, −
2
2
−
1
2
, −
3
2
(𝒄𝒐𝒔(𝜽) , 𝒔𝒊𝒏(𝜽))
By: ferrantetutoring
