UNIT CIRCLE TRIGONOMETRY

The Unit Circle is the circle centered at the origin with radius 1 unit (hence, the “unit” circle). The equation of this circle is x 2 + y 2 = 1 . A diagram of the unit circle is shown below: y 1

x2 + y2 = 1 x -2

-1 -1

1

2

-2

We have previously applied trigonometry to triangles that were drawn with no reference to any coordinate system. Because the radius of the unit circle is 1, we will see that it provides a convenient framework within which we can apply trigonometry to the coordinate plane.

Drawing Angles in Standard Position We will first learn how angles are drawn within the coordinate plane. An angle is said to be in standard position if the vertex of the angle is at (0, 0) and the initial side of the angle lies along the positive x-axis. If the angle measure is positive, then the angle has been created by a counterclockwise rotation from the initial to the terminal side. If the angle measure is negative, then the angle has been created by a clockwise rotation from the initial to the terminal side.

θ

in standard position, where y Terminal side

θ

is positive:

θ

in standard position, where y

θ

is negative:

θ
Initial side

Initial side x

θ

x

Terminal side

Unit Circle Trigonometry

Drawing Angles in Standard Position

Examples The following angles are drawn in standard position: 1. θ = 40 y 2. θ = 160 θ y

x

θ

x

y

3. θ = −320

θ

x

Notice that the terminal sides in examples 1 and 3 are in the same position, but they do not represent the same angle (because the amount and direction of the rotation in each is different). Such angles are said to be coterminal.

Exercises Sketch each of the following angles in standard position. (Do not use a protractor; just draw a brief sketch.)

1. θ = 120 y 2. θ = − 45 y 3. θ = −130 y x

x

x

4. θ = 270 y θ = −90 y 6. θ = 750 y x

x

x

Unit Circle Trigonometry

Drawing Angles in Standard Position

Labeling Special Angles on the Unit Circle

We are going to deal primarily with special angles around the unit circle, namely the multiples of 30o, 45o, 60o, and 90o. All angles throughout this unit will be drawn in standard position. First, we will draw a unit circle and label the angles that are multiples of 90o. These angles, known as quadrantal angles, have their terminal side on either the x-axis or the yaxis. (We have limited our diagram to the quadrantal angles from 0o to 360o.)
Multiples of 90o: y
90o
1

180o
-1 1

0o 360o

x

Note: The 0o angle is said to be coterminal with the 360o angle. (Coterminal angles are angles drawn in standard position that share a terminal side.)

-1

270o

Next, we will repeat the same process for multiples of 30o, 45o, and 60o. (Notice that there is a great deal of overlap between the diagrams.)
Multiples of 30o: y
90o
1

Multiples of 45o: y
90o

120o 150o 180o
-1

60o 30o
1

135

o

1

45o 0o 360o

0o 360o

x

180o
-1 1

x

210o 240o
-1

330o 270 o 300o

225o

-1

315o o 270

Multiples of 60o: y

120o

1

60o

180o
-1 1

0o 360o

x

240o

-1

300o

Unit Circle Trigonometry

Labeling Special Angles on the Unit Circle

Putting it all together, we obtain the following unit circle with all special angles labeled: y 90o 120o 135o 150o
1

60o 45o 30o

180o
-1 1

0o 360o

x

210o 225o 240o
-1

330o 315o 300 o 270o

Coordinates of Quadrantal Angles and First Quadrant Angles

We want to find the coordinates of the points where the terminal side of each of the quadrantal angles intersects the unit circle. Since the unit circle has radius 1, these coordinates are easy to identify; they are listed in the table below. y 90o 1

(0, 1)

(-1, 0)
180o -1 1

0o

(1, 0)

360o

x

-1 270o

Angle 0o 90o 180o 270o 360o

Coordinates (1, 0) (0, 1) (-1, 0) (0, -1) (1, 0)

(0, -1)

We will now look at the first quadrant and find the coordinates where the terminal side of the 30o, 45o, and 60o angles intersects the unit circle.

Unit Circle Trigonometry

Coordinates of Quadrantal Angles and First Quadrant Special Angles

First, we will draw a right triangle that is based on a 30o reference angle. (When an angle is drawn in standard position, its reference angle is the positive acute angle measured from the x-axis to the angle’s terminal side. The concept of a reference angle is crucial when working with angles in other quadrants and will be discussed in detail later in this unit.) y 1

30o
-1 1

x

-1

Notice that the above triangle is a 30o-60o -90o triangle. Since the radius of the unit circle is 1, the hypotenuse of the triangle has length 1. Let us call the horizontal side of the triangle x, and the vertical side of the triangle y, as shown below. (Only the first quadrant is shown, since the triangle is located in the first quadrant.) y 1

( x, y )
1
30o 60o

y x
1

x

We want to find the values of x and y, so that we can ultimately find the coordinates of the point ( x, y ) where the terminal side of the 30o angle intersects the unit circle. Recall our theorem about 30o-60o-90o triangles: In a 30o-60o-90o triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the longer leg is 3 times the length of the shorter leg. Unit Circle Trigonometry
Coordinates of Quadrantal Angles and First Quadrant Special Angles

Since the length of the hypotenuse is 1 and it is twice the length of the shorter leg, y, we can say that y = 1 . Since the longer leg, x, is 3 times the length of the shorter leg, we 2 can say that x =
1 2

3 , or equivalently, x =

3 2

.

Based on the values of the sides of the triangle, we now know the coordinates of the point ( x, y ) where the terminal side of the 30o angle intersects the unit circle. This is the point

(

3 2

, 1 , as shown below. 2 y )

1

(
1
30 o 3 2

,1 2

) x 60o

1 2
1

3 2

We will now repeat this process for a 60o reference angle. We first draw a right triangle that is based on a 60o reference angle, as shown below. y 1

( x, y )

1
60o

y x

x

1

We again want to find the values of x and y. The triangle is a 30o-60o -90o triangle. Since the length of the hypotenuse is 1 and it is twice the length of the shorter leg, x, we can say that x = 1 . Since the longer leg, y, is 3 times the length of the shorter leg, we can say 2 that y =
1 2

3 , or equivalently, y =

3 2

.

Unit Circle Trigonometry

Coordinates of Quadrantal Angles and First Quadrant Special Angles

Based on the values of the sides of the triangle, we now know the coordinates of the point ( x, y ) where the terminal side of the 60o angle intersects the unit circle. This is the point

(

1 2

,

3 2

) , as shown below.

y

1

(
1
60o
1 2

1 2

,

3 2

)

30o
3 2

1

x

We will now repeat this process for a 45o reference angle. We first draw a right triangle that is based on a 45o reference angle, as shown below. y 1

( x, y )
1 y
45o
1

x

x

This triangle is a 45o-45o-90o triangle. We again want to find the values of x and y. Recall our theorem from the previous unit: In a 45o-45o-90o triangle, the legs are congruent, and the length of the hypotenuse is times the length of either leg.
2

Since the length of the hypotenuse is 2 times the length of either leg, we can say that the hypotenuse has length x 2 . But we know already that the hypotenuse has length 1, so x 2 = 1 . Solving for x, we find that x = 12 . Rationalizing the denominator, x= 1 2

⋅

2 2

=

2 2

. Since the legs are congruent, x = y , so y =

2 2

.

Unit Circle Trigonometry

Coordinates of Quadrantal Angles and First Quadrant Special Angles

Based on the values of the sides of the triangle, we now know the coordinates of the point ( x, y ) where the terminal side of the 45o angle intersects the unit circle. This is the point

(

2 2

,

2 2

) , as shown below.

y

1

(
1
45o
2 2

2 2

,

2 2

) x 45o
2 2
1

Putting together all of the information from this section about quadrantal angles as well as special angles in the first quadrant, we obtain the following diagram: y ( 0, 1)
90o
1

(

1 2

, o 3 2

)

60

45

o

(

2 2

,

2 2 3 2

)
,1 2

30o

(
0o

) x ( −1, 0 )

180o
-1 1

(1, 0 ) o 360

-1

( 0,

270o

− 1)

We will use these coordinates in later sections to find trigonometric functions of special angles on the unit circle.

Unit Circle Trigonometry

Coordinates of Quadrantal Angles and First Quadrant Special Angles

Definitions of the Six Trigonometric Functions

We will soon learn how to apply the coordinates of the unit circle to find trigonometric functions, but we want to preface this discussion with a more general definition of the six trigonometric functions.

Definitions of the Six Trigonometric Functions: General Case Let θ be an angle drawn in standard position, and let P ( x, y ) represent the point where the terminal side of the angle intersects the circle x 2 + y 2 = r 2 . The six trigonometric functions are defined as follows:

sin (θ ) = cos (θ ) = tan (θ ) =

y r x r y x

csc (θ ) = sec (θ ) =

1 r = sin (θ ) y 1 r = cos (θ ) x
1 x = tan (θ ) y

( y ≠ 0) ( x ≠ 0)
( y ≠ 0)

( x ≠ 0)

cot (θ ) =

Alternative definitions for the tangent and the cotangent functions are as follows:

tan (θ ) =

cos (θ )

sin (θ )

cot (θ ) =

cos (θ ) sin (θ )

The above functions are not really new to us if we relate them to our previous unit on right triangle trigonometry. For the purpose of remembering the formulas, we will choose to draw an angle θ in standard position in the first quadrant, and then draw a right triangle in the first quadrant which contains that angle, inscribed in the circle x 2 + y 2 = r 2 . (Remember that the circle x 2 + y 2 = r 2 is centered at the origin with radius r.) We label the horizontal side of the triangle x, the vertical side y, and the hypotenuse r (since it represents the radius of the circle.) A diagram of the triangle is shown below.

Unit Circle Trigonometry

Definitions of the Six Trigonometric Functions

y

x2 + y 2 = r 2 r θ

y x x

Notice that the formula x 2 + y 2 = r 2 (the equation of the circle) is simply the Pythagorean Theorem as it relates to the sides of the triangle. Recall the formulas for the basic trigonometric ratios which we learned in the previous unit on right triangle trigonometry, shown below in abbreviated form:

Mnemonic: SOH-CAH-TOA sin(θ ) =
Opposite Hypotenuse

cos(θ ) =

Adjacent Hypotenuse

tan(θ ) =

Opposite Adjacent

and the reciprocal trigonometric ratios:

csc(θ ) =

1 sin(θ )

=

Hypotenuse Opposite

sec(θ ) =

1 cos(θ )

=

Hypotenuse Adjacent

cot(θ ) =

1 tan(θ )

=

Adjacent Opposite

Using these formulas in the triangle from the diagram above, we obtain our six trigonometric functions which we apply to the coordinate plane: sin(θ ) = cos(θ ) = tan(θ ) = = = y x

Opposite Hypotenuse

y r

csc(θ ) = sec(θ ) = cot(θ ) =

1 sin(θ )

= = =

Hypotenuse Opposite

= = x y

r y

Adjacent Hypotenuse

x r

1 cos(θ )

Hypotenuse Adjacent

r x

Opposite Adjacent

=

1 tan(θ )

Adjacent Opposite

=

Unit Circle Trigonometry

Definitions of the Six Trigonometric Functions

Note that even though we drew the right triangle in the first quadrant in order to easily relate these formulas to right triangle trigonometry, these definitions apply to any angle. (We will discuss later how to properly draw right triangles in other quadrants. Right triangles can not be drawn to illustrate the quadrantal angles, but the above formulas still apply.)

Finally, let us justify the new formulas for tangent and cotangent. (We will do this algebraically, not with our right triangle picture from above.) Our alternative definition sin θ y x for the tangent ratio is tan (θ ) = cos((θ )) . If we substitute sin(θ ) = r and cos(θ ) = r into this equation, we can see that our alternative definition is equivalent to our definition y tan(θ ) = x . The algebraic justification is shown below: sin (θ ) y y r y = r = ⋅ = x r x x r cos (θ ) sin (θ )

tan (θ ) =

cos (θ )

Similarly, our alternative definition for the cotangent ratio is cot (θ ) =

. If we

y x substitute sin(θ ) = r and cos(θ ) = r into this equation, we can see that our alternative 1 definition for cotangent is equivalent to the formula cot(θ ) = tan(θ ) = x . The algebraic y

justification is shown below: cos (θ ) sin (θ ) x x r x = r = ⋅ = y r y y r

cot (θ ) =

Let us now apply our trigonometric definitions to the unit circle. Since the unit circle has radius 1, we can see that our trigonometric functions are greatly simplified: sin(θ ) = cos (θ ) = tan (θ ) = y y = =y r 1 x x = =x r 1

csc (θ ) = sec (θ ) = cot (θ ) =

r 1 = , y≠0 y y r 1 = , x≠0 x x

y sin (θ ) , x≠0 = x cos (θ )

x cos (θ ) , y≠0 = y sin (θ )

Unit Circle Trigonometry

Definitions of the Six Trigonometric Functions

Definitions of the Six Trigonometric Functions: Special Case of the Unit Circle Let θ be an angle drawn in standard position, and let P ( x, y ) represent the point where the terminal side of the angle intersects the unit circle x 2 + y 2 = 1 . The six trigonometric functions are defined as follows: sin (θ ) = y cos (θ ) = x csc (θ ) = sec (θ ) = 1 1 = sin (θ ) y 1 1 = cos (θ ) x

( y ≠ 0)

( x ≠ 0)
( y ≠ 0)

tan (θ ) =

y sin (θ ) = x cos (θ )

( x ≠ 0)

cot (θ ) =

1 x cos (θ ) = = tan (θ ) y sin (θ )

Note that these definitions apply ONLY to the unit circle!

Note that the value of a trigonometric function for any given angle remains constant regardless of the radius of the circle. For example, let us suppose that we want to find sin ( 60 ) in two different ways, using the following two diagrams. (Only the first quadrant is shown, since that is all that is needed for the problem.) Diagram 1: y Find sin 60 . Solution: Since the circle shown has radius 12, we must use the general definition for the sine function, y sin (θ ) = . Since the point given is r 6, 6 3 , the y value is 6 3 . We know from

( )

12

P 6, 6 3

(

)

(

)

60o
12

x

the diagram that r = 12 . Therefore, sin ( 60 ) = y 6 3 3 = = . r 12 2

Unit Circle Trigonometry

Definitions of the Six Trigonometric Functions

Diagram 2: y Find sin 60 . Solution: Since the circle shown has radius 1, we can use the shortened definition of the sine function sin (θ ) = y . Since the point given is 1 , 23 , the 2

( )

1

(

1 2

,

3 2

)

(

)

y value is
60o
1

3 2

.

x

Therefore, sin 60 = y =

( )

3 2

.

We will use the unit circle (and thus the shortened definitions) to evaluate the trigonometric functions of special angles. (We will see some examples later in the unit where the general definitions of the trigonometric functions can be useful.)

Examples Find the exact values of the following:

1. cos 90 4. 7.

( ) csc (180 ) tan ( 45 )

2. sin 45 5. 8.

( ) sec ( 60 ) sin ( 270 )

3. tan 30 6. 9.

( ) cot ( 0 ) sec ( 30 )

Solutions: First, recall our unit circle with the coordinates which we have filled in so far: y ( 0, 1)
90o
1

(

1 2

,

3 2

60o 45

o

) (

2 2

,

2 2 3 2

30o

(

)
,1 2

) x ( −1, 0 ) 180

o

-1

1

0o 360o

(1, 0 )

-1

( 0,
Unit Circle Trigonometry

270o

− 1)

Definitions of the Six Trigonometric Functions

1. cos 90

( ) ( )

The terminal side of a 90o angle intersects the unit circle at the point ( 0, 1) . Using the definition cos (θ ) = x , we conclude that cos 90 = 0 . 2. sin 45

( )
( )
2 2

The terminal side of a 45o angle intersects the unit circle at the point Using the definition sin (θ ) = y , we conclude that sin 45 = 3. tan 30 .

(

2 2

,

2 2

).

( )
( )
1 2 3 2

The terminal side of a 30o angle intersects the unit circle at the point y Using the definition tan (θ ) = x , we conclude that tan 30 =

(

3 2 2 3

,1 . 2 =
1 3

)

= 1⋅ 2

.

Rationalizing the denominator, tan 30 =

( )

1 3

⋅

3 3

=

3 3

.

4. csc 180

(

)
1 sin (θ )

The terminal side of a 180o angle intersects the unit circle at the point ( −1, 0 ) . Using the definition csc (θ ) = undefined, since 5. sec 60
1 0

= 1 , we conclude that csc 180 y

(

) is

is undefined.

( )
( )
1
1 2

The terminal side of a 60o angle intersects the unit circle at the point Using the definition sec (θ ) = 1 , we conclude that sec 60 = x 6. cot 0

(

1 2

,

3 2

).

2 = 1⋅ 1 = 2 .

( ) ( ) is undefined, since

The terminal side of the 0o angle intersects the unit circle at the point (1, 0 ) . Using the definition cot (θ ) = x , we conclude that cot 0 y
1 0

is undefined.

Unit Circle Trigonometry

Definitions of the Six Trigonometric Functions

7. tan 45

( )
2 2 2 2

The terminal side of a 45o angle intersects the unit circle at the point y Using the definition tan (θ ) = x , we conclude that tan ( 45 ) =

(
2 2

2 2

,

2 2

).

=

⋅

2 2

=1.

8. sin ( 270

)

The terminal side of a 270o angle intersects the unit circle at the point ( 0, − 1) . Using the definition sin (θ ) = y , we conclude that sin 270 = −1 . 9. sec 30

(

)

( )
1 cos (θ )

The terminal side of a 30o angle intersects the unit circle at the point Using the definition sec (θ ) = sec 30 =

(

3 2

,1 . 2

)

= 1 , we conclude that x

( )

1
3 2 2 3

= 1⋅

2 3

=

2 3

. Rationalizing the denominator,

sec ( 30 ) =

⋅

3 3

=

2 3 3

.

Example Find tan 90

( ) and cot ( 90 ) .

Solution: The terminal side of a 90o angle intersects the unit circle at the point ( 0, 1) . To find tan 90 , we use the definition tan (θ ) = is undefined, since
1 0

( )

y x

and conclude that tan 90

( )

is undefined. x y

To find cot 90 , we use the definition cot (θ ) =
0 cot 90 = 1 = 0 .

( )

( )

and conclude that

Important Note: If we want to find cot 90 cot (θ ) =
1 undefined

1 tan (θ )

, knowing that tan 90

( ) is undefined, we cannot compute

( ) and we instead use the definition x y

. In this case, we need to instead use the direct formula cot (θ ) =

as

we have done above.

Unit Circle Trigonometry

Definitions of the Six Trigonometric Functions

Coordinates of All Special Angles on the Unit Circle

We will now label the coordinates of all of the special angles in the unit circle so that we can apply the trigonometric functions to angles in any quadrant. First, recall our unit circle with the coordinates we have filled in so far. y ( 0, 1)
90o
1

(

1 2

,

3 2

60o 45

o

) (

2 2

,

2 2 3 2

30o

(

)
,1 2

) x ( −1, 0 ) 180

o

-1

1

0o 360o

(1, 0 )

-1

( 0,

270o

− 1)

We will now draw all of the angles (between 0o and 360o) that have a 30o reference angle.

Definition of Reference Angle When an angle is drawn in standard position, its reference angle is the positive acute angle measured from x-axis to the angle’s terminal side.

Angles with Reference angles of 30o: Points A, B, C, and D each represent the point where the terminal side of an angle intersects the unit circle. For each of these points, we wish to determine the angle measure (in standard position from 0o to 360o) as well as the coordinates of that particular angle.
1

y

B 30o 30o 30o 30o

A

-1

1

x

C
-1

D

Unit Circle Trigonometry

Coordinates of All Special Angles on the Unit Circle

Point A is located along the terminal side of a 30o angle. This point should already be familiar to us; the coordinates of point A are 23 , 1 . 2

(

)

Point B is located along the terminal side of a 150o angle, since 180 − 30 = 150 . The coordinates of point B are almost identical to those of point A, except that the x-value is negative in the second quadrant, so the coordinates of point B are − 23 , 1 . 2

(

)

Point C is located along the terminal side of a 210o angle, since 180 + 30 = 210 . The coordinates of point C are almost identical to those of point A, except that the x-value and y-value are negative in the third quadrant, so the coordinates of point C are − 23 , − 1 . 2

(

)

Point D is located along the terminal side of a 330o angle, since 360 − 30 = 330 . The coordinates of point D are almost identical to those of point A, except that the y-value is negative in the fourth quadrant, so the coordinates of point D are 3 1 2 ,− 2 .

(

)

Below is a diagram with all of the 30o reference angles between 0o and 360o, along with their respective coordinates: y 1

(− (−
3 2

3 2

, 1 150 2
-1

)

30
30o 30o 30o 30o

(

3 2

,1 2 x ) )

1

, − 1 210 2
-1

)

330

(

3 2

,−1 2

We have gone into great detail describing the justification for the angles and coordinates related to 30o reference angles. Since the explanations for the 45o and 60o reference angles are so similar, we will skip the details and simply give the final diagrams with the angles and coordinates labeled.

Unit Circle Trigonometry

Coordinates of All Special Angles on the Unit Circle

Angles with Reference angles of 45o: Below is a diagram with all of the 45o reference angles between 0o and 360o, along with their respective coordinates: y (−

2 2

,

2 2

) 135
-1

1

45

(

2 2

,

2 2

)

45o 45o

45o 45o

1

x

(−

2 2

,−

2 2

) 225

-1

315

(

2 2

,−

2 2

)

Angles with Reference angles of 60o: Below is a diagram with all of the 60o reference angles between 0o and 360o, along with their respective coordinates: y (−

1 2

,

3 2

) 120
60o
-1

1

60
60o 60o
1

(

1 2

,

3 2

) x 60o

(−

1 2

,−

3 2

) 240

-1

300

(

1 2

,−

3 2

)

We will now put all of the previous information together in one diagram which includes the coordinates of all special angles on the unit circle.

Unit Circle Trigonometry

Coordinates of All Special Angles on the Unit Circle

Coordinates of All Special Angles on the Unit Circle

y
90 ( 0, 1)
1

(−

2 2

( − , ) 120 , ) 135
1 2 3 2 2 2

60

(

1 2

,

3 2

)

45

(

2 2

,

2 2

)
,1 2

(−

3 2

, 1 150 2

)

30

(

3 2

) x ( −1, 0 ) 180
-1 1

0 (1, 0 )

360o

(−

3 2

, − 1 210 2
2 2

)

330 315

(

3 2

,−1 2
2 2

)

(−

,−

2 2

(−

) 225
1 2

,−

3 2

) 240

-1

270 ( 0, − 1)

300

(

1 2

,−

3 2

( )

2 2

,−

)

Evaluating Trigonometric Functions of Any Special Angle

Now that we know the coordinates of all the special angles on the unit circle, we can use these coordinates to find trigonometric functions of any special angle (i.e. any multiple of 30o, 45o, or 60o). First, let us review the concept of coterminal angles. Definition of Coterminal Angles Coterminal Angles are angles drawn in standard position that share a terminal side. For any angle θ , an angle coterminal with θ can be obtained by using the formula θ + k ⋅ 360 , where k is any integer.

(

)

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Any Special Angle

We already have discussed the fact that a 0o angle is coterminal with a 360o angle. Below are more examples of coterminal angles.

Example Find four angles that are coterminal with a 30o angle drawn in standard position.

Solution: To find angles coterminal with 30o, we add to it any multiple of 360o, since adding 360o means adding exactly one revolution, which keeps the terminal side of the angle in the same position. There are an infinite number of coterminal angles for any given angle, so the following solutions are not unique. 30 + 1⋅ 360 = 390 30 30 30

( ) + 2 ⋅ ( 360 ) = 750 + ( −1) ⋅ ( 360 ) = −330 + ( −2 ) ⋅ ( 360 ) = −690

Therefore, four angles coterminal with 30o are 390o, 750o, -330o, and -690o.

Since coterminal angles share the same terminal side, they intersect the same point on the unit circle. Therefore, since 30o intersects the unit circle at the point 23 , 1 , the angles 2 390o, 750o, -330o, and -690o

( also intersect the unit circle at the point (

3 2

,1 2

) ) . This allows

us to find trigonometric functions of special angles other than those that we have drawn on our unit circle between 0o and 360o. Next, recall the definitions of the trigonometric functions as they apply to the unit circle: sin (θ ) = y cos (θ ) = x y tan (θ ) = x = cos (θ ) sin (θ )

csc (θ ) = sec (θ ) =

1 sin (θ )

= =

1 y

( y ≠ 0)
( x ≠ 0) cos (θ ) sin (θ )

1 cos(θ )
1 tan (θ )

1 x

( x ≠ 0)

cot (θ ) =

=x= y

( y ≠ 0)

We can now use these definitions to find trigonometric functions of any special angle.

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Any Special Angle

Examples

Find the exact values of the following: 1. sin 240 4. 7.

( ) sec ( 30 ) sin ( −240 )

2. cos 315 5. 8.

( ) tan (135 ) tan ( −150 )

3. cot 210 6. 9.

( ) csc ( 630 ) sec ( 810 )

Solutions: 1. sin 240

(

)

The terminal side of a 240o angle intersects the unit circle at the point − 1 , − 23 . Using the definition sin (θ ) = y , we conclude that 2

(

sin 240 = − 2. cos 315

(

)

)

3 2

.

( (

)

The terminal side of a 315o angle intersects the unit circle at the point 2 2 2 , − 2 . Using the definition cos (θ ) = x , we conclude that

(

)

cos 315 = 3. cot 210

)

2 2

.

(
(

)

The terminal side of a 210o angle intersects the unit circle at the point − 23 , − 1 . Using the definition cot (θ ) = x , we conclude that 2 y

(

)

cot 210 =

)

−

3 2 −1 2

=

− 3 2

⋅ −12 = 3 . .

4. sec 30

( )
1 cos(θ )

The terminal side of a 30o angle intersects the unit circle at the point Using the definition sec (θ ) =

(

3 2

,1 . 2

)

= 1 , we conclude that x

( ) sec ( 30 ) = sec 30 =

1
3 2

= 1⋅

2 3

=
2 3 3

2 3

. Rationalizing the denominator,

2 3

⋅

3 3

=

.

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Any Special Angle

5. tan 135

(

)

The terminal side of a 135o angle intersects the unit circle at the point y − 22 , 22 . Using the definition tan (θ ) = x , we conclude that

(

)

tan 135 =

(

)

2 2

−

2 2

=

2 2

⋅ −2 = −1 . 2

6. csc ( 630

)
1 sin (θ )

A 630o angle is coterminal with a 270o angle (since 630 − 360 = 270 ) and its terminal side intersects the unit circle at the point ( 0, − 1) . Using the definition csc (θ ) = 7. sin ( −240 o = 1 , we conclude that csc 630 = y

(

)

1 −1

= −1 .

)

A -240 angle is coterminal with a 120o angle (since −240 + 360 = 120 ) and its terminal side intersects the unit circle at the point − 1 , 23 . Using the 2 definition sin (θ ) = y , we conclude that sin −240 = 8. tan ( −150

(

)

(

)

3 2

.

)

A -150o angle is coterminal with a 210o angle (since −150 + 360 = 210 ) and its terminal side intersects the unit circle at the point − 23 , − 1 . Using the 2

(

)

y definition tan (θ ) = x , we conclude that tan −150 =

(

)

Rationalizing the denominator, we obtain tan ( −150 ) = 9. sec ( 810

−1 2 = − 23
1 3

−1 2

⋅ −2 = 3
3 3

1 3

.

⋅

3 3

=

)
= 1 , we conclude that sec 810 x

A 810o angle is coterminal with a 90o angle (since 810 − 2 ⋅ 360 = 90 ) and its terminal side intersects the unit circle at the point ( 0, 1) . Using the definition sec (θ ) =
1 0 1 cos (θ )

(

)

(

) is undefined, since

is undefined.

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Any Special Angle

Exercises

Find the exact values of the following: 1. cos (150 4. cot 60 7.

( ) cot ( −450 )

)

2. tan ( 330 5. 8.

) csc ( 225 ) cos ( 495 )

3. sin ( 225 9.

6. tan −120

( ) sin ( 3630 )

)

Methods of Finding Trigonometric Functions of Special Angles

We can see that the unit circle assists us greatly in finding trigonometric functions of special angles. But what if we don’t have a diagram of the unit circle, and we don’t wish to draw one? It is assumed that the quadrantal angles (multiples of 90o -- on the axes) are fairly easy to visualize without a diagram. One option is to memorize the coordinates of the special angles in the first quadrant of the unit circle (30o, 45o, and 60o), and use those values to find the trigonometric functions of angles in other quadrants. We will discuss a few ways of finding the basic trigonometric functions of 30o, 45o, and 60o (other than memorizing the coordinates of these special angles in the first quadrant of the unit circle). One method is to use a basic 45o-45o-90o triangle and a 30o-60o-90o triangle to derive the trigonometric functions of 30o, 45o, and 60o. We learned earlier in the unit that the trigonometric functions are constant for any given angle; for example, sin 30 is always

( )

, regardless of the size of the radius of the circle – or the size of the triangle drawn. So we can choose any 45o-45o-90o triangle and 30o-60o-90o triangle to work from. For simplicity, let us choose both triangles to have a shorter leg of length 1, as shown below:

1 2

45o

60o

1
45o 30o

1

Recall that in a 45o-45o-90o triangle, the legs are congruent, and the length of the hypotenuse is 2 times the length of either leg. In a 30o-60o-90o triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the longer leg is 3 times the length of the shorter leg.

Unit Circle Trigonometry

Methods of Finding Trigonometric Functions of Special Angles

Filling in the missing sides from the diagram above, we obtain the following triangles:

2
45o

45o

2 1
30o

60o

1

1

3

Now recall the trigonometric ratios that we learned for right triangles (shown below in abbreviated form):

SOH-CAH-TOA sin(θ ) =
Opposite Hypotenuse

cos(θ ) =

Adjacent Hypotenuse

tan(θ ) =

Opposite Adjacent

Using these ratios on the triangles above, we obtain the following: sin(45 ) = cos(45 ) = tan(45 ) =
Opposite Hypotenuse
Adjacent Hypotenuse Opposite Adjacent

= =

1 2
1 2

= =

1 2
1 2

⋅ ⋅

2 2
2 2

= =

2 2
2 2

(Note: This is the y-coordinate of 45o on the unit circle.) (Note: This is the x-coordinate of 45o on the unit circle.)

= 1 =1 1

sin(30 ) =

Opposite Hypotenuse Adjacent Hypotenuse

=

1 2 3 2

(Note: This is the y-coordinate of 30o on the unit circle.) (Note: This is the x-coordinate of 30o on the unit circle.)
1 3

cos(30 ) = tan(30 ) =

=
1 3

Opposite Adjacent

=

=

⋅

3 3

=

3 3

sin(60 ) = cos(60 ) =

Opposite Hypotenuse Adjacent Hypotenuse Opposite Adjacent

=
=
3 1

3 2 1 2

(Note: This is the y-coordinate of 60o on the unit circle.) (Note: This is the x-coordinate of 60o on the unit circle.)

tan(60 ) =

=

The reciprocal trigonometric ratios cosecant, secant, and cotangent can be obtained by simply taking the reciprocals of the sine, cosine, and tangent ratios above. Unit Circle Trigonometry Methods of Finding Trigonometric Functions of Special Angles

Another method of remembering the trigonometric functions of 30o, 45o, and 60o, along with 0o and 90o, is shown below. (This is more of a ‘trick’ for remembering the ratios, so no mathematical justification is given.) Step 1: Label the columns at the top of the chart for all of the special angles between 0o and 90o in ascending order, as shown. Also label the rows with the words “sine” and “cosine.”
0o sine cosine 30o 45o 60o 90o

Step 2: This chart is not yet complete! Write the numbers 0, 1, 2, 3, 4 in the “sine” row, and the numbers 4, 3, 2, 1, 0 in the “cosine” row, as shown.
0o sine cosine 30o 45o 60o 90o

0 4

1 3

2 2

3 1

4 0

This chart is not yet complete!

This chart is not yet complete!

Step 3: This chart is correct, but not yet simplified… For each of the numbers in the “sine” and “cosine” rows, take the square root of the number and then divide by 2.
0o sine cosine 30o 45o 60o 90o

0 2 4 2

1 2 3 2

2 2 2 2

3 2 1 2

4 2 0 2

We then simplify each of the numbers in the chart above. Unit Circle Trigonometry Methods of Finding Trigonometric Functions of Special Angles

Step 4: This is the final version of the chart! Simplify each of the values from the table above, and we obtain our final chart:
0o sine cosine 30o 45o 60o 90o

0

1 2

2 2 2 2

1

3 2

3 2 1 2

1
0

y-values on the unit circle x-values on the unit circle

Examples

Use the methods learned in this section (special right triangles or the chart above) to find the exact values of the following trigonometric functions. Note that these examples were also included in the previous section, but will now be solved using a different method. 1. sin ( 240 4. sec ( 30

)

)

2. cos ( 315 5.

) tan (135 )

3. cot ( 210

)

Solutions: 1. sin 240

(

)
( ) ( ) must be negative. Since

The terminal side of a 240o angle measures 60o to the x-axis, so we use a 60o reference angle. Using either a 30o-60o-90o triangle or our chart from above, we find that sin 60 = 23 . Since a 240o angle is in the third quadrant, and yvalues in the third quadrant are negative, sin 240 sin 60 = 2. cos 315

( )
(

3 2

, we conclude that sin 240 = −

(

)

3 2

.

) ( )
2 2

The terminal side of a 315o angle measures 45o to the x-axis, so we use a 45o reference angle. Using either a 45o-45o-90o triangle or our chart from above, we find that cos 45 = 22 . Since a 315o angle is in the fourth quadrant, and x-values in the fourth quadrant are positive, cos 315 Since cos 45 =

( )

, we conclude that cos 315 =

(

(

)

) must be positive.
.

2 2

Unit Circle Trigonometry

Methods of Finding Trigonometric Functions of Special Angles

3. cot ( 210

)

The terminal side of a 210o angle measures 30o to the x-axis, so we use a 30o cos θ reference angle. Remember that cot (θ ) = tan1(θ ) = sin ((θ )) . If we use the 30o-60o90o triangle, we find that tan 30 = instead use our chart from above, we find that cos ( 30 ) = sin ( 30 ) = 1 , so cot 30 = 2

( )

1 3

, so cot ( 30 ) =

1 tan (θ )

= 3 . If we

3 2

and

( )

cos 30

( ) = sin ( 30 )

3 2 1 2

=

3 2

2 ⋅ 1 = 3 . Since a 210o angle is in

the third quadrant, and both the x and y-values in the third quadrant are negative, this means that both cos ( 210 ) and sin ( 210 ) are negative, therefore their quotient cot ( 210 ) = we conclude that cot ( 210 ) = 3 . 4. sec 30 cos 210 sin 210

( (

) is positive. Since cot ( 30 ) = 3 , )

( )

The terminal side of a 30o angle measures 30o to the x-axis, so we use a 30o reference angle. (A reference angle is not really necessary in this case, since any angle which is located in the first quadrant can be used as-is.) Remember that sec (θ ) = cos1(θ ) . Using either a 30o-60o-90o triangle or our chart from above, we find that cos 30 =

( )

3 2

, so sec ( 30 ) =
2 3

1 cos 30

( )
2 3 3

=

1
3 2

= 1⋅

2 3

=

2 3

.

Rationalizing the denominator, sec ( 30 ) = 5. tan (135

⋅

3 3

=

.

)

The terminal side of a 135o angle measures 45o to the x-axis, so we use a 45o sin θ reference angle. Remember that tan (θ ) = cos((θ )) . If we use the 45o-45o-90o triangle, we find that tan 45 = 1 = 1 . If we instead use our chart from above, 1 sin ( 45 ) =
2 2

( )

and sin ( 45 ) =

2 2

, so tan ( 45 ) =

( ) = cos ( 45 ) sin 45

2 2 2 2

=

2 2

⋅

2 2

= 1 . Since

( ) ( therefore their quotient tan (135 ) = ( conclude that tan (135 ) = −1 .
Unit Circle Trigonometry

a 135o angle is in the second quadrant, the x-value is negative and the y-value is positive, which means that cos 135 is negative and sin 135 is positive;

sin 135

cos 135

) )

) is negative. Since tan ( 45 ) = 1 , we

(

Methods of Finding Trigonometric Functions of Special Angles

Evaluating Trigonometric Functions of Other Angles

Now that we have explored the trigonometric functions of special angles, we will briefly look at how to find trigonometric functions of other angles. First, we will derive an important trigonometric identity, known as a Pythagorean Identity. It can be related to the Pythagorean Theorem; consider the following right triangle (drawn in the first quadrant for simplicity). y x2 + y 2 = r 2 r θ

y x x

We can see that the equation x 2 + y 2 = r 2 (the equation of a circle with radius r) is also the Pythagorean Theorem as it relates to the right triangle above. Let us now direct our attention to the unit circle. Since the radius is 1, any point on the circle itself satisfies the equation x 2 + y 2 = 1 (the equation of a circle with radius 1). On the unit circle, we know that x = cos (θ ) and y = sin (θ ) . Substituting these into the equation x 2 + y 2 = 1 , we obtain the equation

( cos (θ ) ) + ( sin (θ ) )
2

2

= 1 . It is standard practice in trigonometry to write these
2

individual terms in shortened form. We write ( cos (θ ) ) as cos 2 (θ ) , and we write

( sin (θ ) )

2

as sin 2 (θ ) . We then obtain the following trigonometric identity: cos 2 (θ ) + sin 2 (θ ) = 1

The identity cos 2 (θ ) + sin 2 (θ ) = 1 applies to an angle drawn in any circle of radius r, not just the unit circle. A short justification is shown below.
Unit Circle Trigonometry Evaluating Trigonometric Functions of Other Angles

We begin with the general equation of a circle of radius r: x2 + y 2 = r 2 Dividing both sides by r 2 , we obtain the equation x2 y2 r 2 + = . r2 r2 r2 Simplifying the equation, x2 y2 + =1. r2 r2 The equation can then be rewritten as

⎛x⎞ ⎛ y⎞ ⎜ ⎟ + ⎜ ⎟ = 1. ⎝r⎠ ⎝r⎠
For a circle of radius r, we know from our general trigonometric definitions that x y cos (θ ) = and sin (θ ) = . We substitute these into the above equation and r r conclude that cos 2 (θ ) + sin 2 (θ ) = 1 .

2

2

We will now use this Pythagorean identity to help us to find trigonometric functions of angles.

Example If sin (θ ) =

3 5

and 90 < θ < 180 , find the exact values of cos (θ ) and tan (θ ) .

Solution: Since 90 < θ < 180 , the terminal side of θ is in the second quadrant. Looking at 3 our unit circle, we can easily see that there is no special angle with a y-value of 5 . There are two methods by which we can solve this problem, both of which are shown below. Method 1: Our first method is to use the Pythagorean identity cos 2 (θ ) + sin 2 (θ ) = 1 .
3 Since sin (θ ) = 5 , we plug it into the equation

cos 2 (θ ) + sin 2 (θ ) = 1 .
3 cos 2 (θ ) + ( 5 ) = 1 2

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Other Angles

We then simplify the equation and solve for cos (θ ) :
9 cos 2 (θ ) + 25 = 1 9 cos 2 (θ ) = 1 − 25

cos 2 (θ ) = cos (θ ) =

cos 2 (θ ) = 16 25
4 5

25 25

9 − 25

4 OR cos (θ ) = − 5

(We need to choose; see below...)

4 cos (θ ) = − 5

The terminal side of θ is in the second quadrant. We know that x values in the second quadrant are negative, therefore cos (θ ) is negative.

We now want to find tan (θ ) . Using the definition tan (θ ) = tan (θ ) = sin (θ ) cos (θ )

cos (θ )

sin (θ )

,

=

3 5 4 −5

3 = 5 ⋅− 5 = − 3 . 4 4

tan (θ ) = − 3 4

4 We conclude that cos (θ ) = − 5 and tan (θ ) = − 3 . 4

Method 2: Our second method is to draw a right triangle in a circle of radius r. Since 90 < θ < 180 , we draw the right triangle in the second quadrant, as shown below. (When drawing the right triangle, we must make sure that the right triangle has one leg on the x-axis and that one acute angle of the triangle -- the reference angle of θ -has its vertex at the origin.) y x2 + y 2 = r 2

( x, y ) P r y x θ x

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Other Angles

3 Since sin (θ ) = 5 , and we know that sin (θ ) =

y r

, we can label the

diagram with the values y = 3 and r = 5 , as shown below. (The y opposite formula sin (θ ) = r can be remembered by using the ratio hypotenuse , using the reference angle of θ which is inside the triangle.) Note that when choosing how to label the sides of the triangle, the radius of the circle is always chosen to be positive. y x2 + y 2 = r 2

5 3 x θ x

To find the value of x, we then use the equation for the circle (or equivalently, the Pythagorean Theorem) x 2 + y 2 = r 2 : x 2 + 32 = 52 x 2 + 9 = 25 x 2 = 16 x = 4 OR x = − 4 (We need to choose; see below.)

x = −4

The terminal side of θ is in the second quadrant. We know that x values in the second quadrant are negative, therefore we choose the negative value for x.

x To find cos (θ ) , we now use the formula cos (θ ) = r . (The

formula cos (θ ) =

x r

can be remembered by using the ratio

adjacent hypotenuse

,

using the reference angle of θ which is inside the triangle.) Note that we can NOT use the formula cos (θ ) = x , since that trigonometric definition only applies to the unit circle, and the circle that we created has radius 5. x cos (θ ) = r =
−4 5

4 =−5

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Other Angles

y To find tan (θ ) , we now use the formula tan (θ ) = x . (Remember

that the formula tan (θ ) = opposite adjacent

y x

can be remembered by using the ratio

, using the reference angle of θ which is inside the triangle.)
3 −4

y tan (θ ) = x =

=−3 4

Note that we could have also used the formula tan (θ ) = sin (θ ) cos (θ )

=

3 5 −4 5

3 3 = 5 ⋅ −54 = − 4

4 We conclude that cos (θ ) = − 5 and tan (θ ) = − 3 . 4

We now wish to derive two other Pythagorean identities. Both identities can be easily derived from the identity cos 2 (θ ) + sin 2 (θ ) = 1 . First, begin with the identity cos 2 (θ ) + sin 2 (θ ) = 1 . Now divide each term by cos 2 (θ ) : cos 2 (θ ) cos (θ )
2

+

cos (θ )
2

sin 2 (θ )

=

cos 2 (θ )

1

Since

sin (θ ) cos (θ )

= tan (θ ) and

1 cos (θ )

= sec (θ ) , we can simplify the above equation and

obtain the following Pythagorean identity:
1 + tan 2 (θ ) = sec 2 (θ )

In a similar fashion, we again begin with the identity cos 2 (θ ) + sin 2 (θ ) = 1 . Now divide each term by sin 2 (θ ) : cos 2 (θ ) sin (θ )
2

+

sin 2 (θ ) sin (θ )
2

=

sin (θ )
2

1

Since

cos (θ ) sin (θ )

= cot (θ ) and

1 sin (θ )

= csc (θ ) , we can simplify the above equation and

obtain the following Pythagorean identity: cot 2 (θ ) + 1 = csc 2 (θ )

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Other Angles

Pythagorean Identities cos 2 (θ ) + sin 2 (θ ) = 1 1 + tan 2 (θ ) = sec 2 (θ ) cot 2 (θ ) + 1 = csc 2 (θ )

Example If cot (θ ) =

11 5

and 90 < θ < 270 , find the exact values of sin (θ ) and cos (θ ) .

Solution: Since 90 < θ < 270 , the terminal side of θ is either in the second or the third quadrant. Recall the definition for cotangent cot (θ ) = x ; we are given that y cot (θ ) = and y is positive, so the quotient cot (θ ) =
11 5

, which is a positive number. In the second quadrant, x is negative x y

is negative; therefore the terminal

side of θ can not lie in the second quadrant. In the third quadrant, on the other hand, both x and y are negative, so the quotient cot (θ ) = x is positive, which is y consistent with the given information. We then conclude that the terminal side of θ is in the third quadrant. There are two methods by which we can solve this problem, both of which are shown below. Method 1: Our first method is to use the Pythagorean identity cot 2 (θ ) + 1 = csc 2 (θ ) . Since cot (θ ) =
2 11 5 2

, we plug it into the equation

cot (θ ) + 1 = csc (θ ) .

( )
11 5

2

+ 1 = csc 2 (θ )

We then simplify the equation and solve for csc (θ ) :

( )
11 5 11 25 11 25 36 25

2

+ 1 = csc 2 (θ )

+ 1 = csc 2 (θ ) + 25 = csc 2 (θ ) 25 = csc 2 (θ )

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Other Angles

csc (θ ) =

6 5

OR csc (θ ) = − 6 5

(We need to choose; see below.)

csc (θ ) = − 6 5

The terminal side of θ is in the third quadrant. We know that y values in the third quadrant are negative, which means that sin (θ ) is negative. Since csc (θ ) = negative.
1 sin (θ )

, we conclude that csc (θ ) is

We now want to find sin (θ ) . Since the cosecant and sine functions are reciprocals of each other, we can rearrange the equation csc (θ ) = sin1(θ ) to say that sin (θ ) = csc1(θ ) . sin (θ ) =
1 csc(θ )

=

1 = 1⋅ −65 = − 5 . 6 6 −5

Our final step is to find cos (θ ) . Using the identity cos 2 (θ ) + sin 2 (θ ) = 1 ,

cos 2 (θ ) + ( − 5 ) = 1 6
2 25 cos 2 (θ ) + 36 = 1 25 cos 2 (θ ) = 1 − 36 = 36 36 25 − 36

cos 2 (θ ) =

11 36 11 6

cos (θ ) =

OR cos (θ ) = −
11 6

11 6

(We need to choose; see below.)

cos (θ ) = −

The terminal side of θ is in the third quadrant. We know that x values in the third quadrant are negative, which means that cos (θ ) is negative.

We conclude that sin (θ ) = − 5 and cos (θ ) = − 6

11 6

.

Method 2: Our second method is to draw a right triangle in a circle of radius r. Since we have determined that the terminal side of θ is in the third quadrant, we draw the right triangle in the third quadrant, as shown below. (When drawing the right triangle, we must make sure that the right triangle has one leg on the x-axis and that one acute angle of the triangle -- the reference angle of θ -- has its vertex at the origin.)

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Other Angles

y

x2 + y 2 = r 2

x y r

θ

x

( x, y ) P
Since cot (θ ) = , and we know that cot (θ ) = x , we can label y

11 5

the diagram with the values x = − 11 and y = −5 , as shown below. We can NOT label the sides of the triangle with the values x = 11 and y = 5 , since the triangle is in the third quadrant, where both x and y are negative. (The formula cot (θ ) = x can be y remembered by using the ratio adjacent opposite

, using the reference angle of

θ which is inside the triangle.) y x2 + y 2 = r 2

− 11

θ

x

-5

r

(−

11, −5 P

)

To find the value of x, we then use the equation for the circle (or equivalently, the Pythagorean Theorem) x 2 + y 2 = r 2 :

( − 11)
36 = r 2 r =6
Unit Circle Trigonometry

2

+ ( −5 ) = r 2 , so 11 + 25 = r 2
2

(Note that the radius of the circle is always positive.)
Evaluating Trigonometric Functions of Other Angles

y To find sin (θ ) , we now use the formula sin (θ ) = r . (The formula

sin (θ ) =

y r

can be remembered by using the ratio

opposite hypotenuse

, using

the reference angle of θ which is inside the triangle.) Note that we can NOT use the formula sin (θ ) = y , since that trigonometric definition only applies to the unit circle, and the circle that we created has radius 6. y sin (θ ) = r =

−5 6

=−5 6

x To find cos (θ ) , we now use the formula cos (θ ) = r . (The

formula cos (θ ) =

x r

can be remembered by using the ratio

adjacent hypotenuse

,

using the reference angle of θ which is inside the triangle.) Note that we can NOT use the formula cos (θ ) = x , since that trigonometric definition only applies to the unit circle, and the circle that we created has radius 6. x cos (θ ) = r = − 11 6

=−

11 6

We conclude that sin (θ ) = − 5 and cos (θ ) = − 6

11 6

.

Exercises Answer the following, using either of the two methods described in this section.
5 1. If cos (θ ) = − 13 and 180 < θ < 360 , find the exact values of sin (θ ) and

tan (θ ) .

2. If csc (θ ) = cot (θ ) .

25 7

and 90 < θ < 270 , find the exact values of cos (θ ) and

3. If tan (θ ) = − sin (θ ) .

5 2

and 180 < θ < 360 , find the exact values of cos (θ ) and

Unit Circle Trigonometry

Evaluating Trigonometric Functions of Other Angles

Graphs of the Sine and Cosine Functions

In this section, we will learn how to graph the sine and cosine functions. To do this, we will once again use the coordinates of the special angles from the unit circle. We will first make a chart of values for y = f ( x) = sin( x) , where x represents the degree measure of the angle. In the column for the y values, the exact value has also been written as a decimal, rounded to the nearest hundredth for graphing purposes.

x

y = sin(x)
1 2 2 2 3 2

x

y = sin(x)

0o 30o 45o 60o 90o 120o 135o 150 o 0 = 0.5
≈ 0.71 ≈ 0.87 1 ≈ 0.87 ≈ 0.71

180o 210o 225o 240o 270o 300o 315o 330 360o o 0 − = −0.5
1 2

− − − −

2 2 3 2

≈ −0.71 ≈ −0.87

3 2 2 2 1 2

3 2 2 2 1 2

−1 ≈ −0.87
≈ −0.71

= 0.5

− = −0.5 0

We now plot the above x and y values on the coordinate plane, as shown:
1.4 y 1.2 1.0 0.8 0.6 0.4 0.2 x

-0.2 -0.4 -0.6 -0.8 -1.0 -1.2 -1.4

Unit Circle Trigonometry

Graphs of the Sine and Cosine Functions

Drawing a smooth curve through the points which we have plotted, we obtain the following graph of y = f ( x) = sin( x) :
1.4 y 1.2 1.0 0.8 0.6 0.4 0.2

y = f ( x) = sin( x)

-0.2 -0.4 -0.6 -0.8 -1.0 -1.2 -1.4

0o

30o

60o

90o

120o 150o 180o 210o 240o 270o 300o 330o 360o

x

Since 360o is coterminal with 0o, their y-values are the same. This is the case with any coterminal angles; 450o is coterminal with 90o, 540o is coterminal with 180o, etc. For this reason, the above graph will repeat itself over and over again, as shown below: y 1

y = f ( x) = sin( x)

-360o -270o -180o

-90o

0o

90o

180o

270o

360o

450o

540o

630o

x 720o

-1

Unit Circle Trigonometry

Graphs of the Sine and Cosine Functions

We will now repeat the same process to graph the cosine function. First, we will make a chart of values for y = f ( x) = cos( x ) , where x represents the degree measure of the angle. In the column for the y values, the exact value has also been written as a decimal, rounded to the nearest hundredth for graphing purposes.

x

y = cos(x)
3 2 2 2 1 2

x

y = cos(x) o 0 30o 45 o o

o

1 ≈ 0.87
≈ 0.71

180 210o 225 o o

− −

3 2 2 2 1 2

−1 ≈ −0.87
≈ −0.71

60 90o 120o 135o 150 o = 0.5 0 − 1 = −0.5 2
− −
2 2 3 2

240 270o 300o 315o 330 360o o − = −0.5 0 1 = 0.5 2
2 2 3 2

≈ −0.71 ≈ −0.87

≈ 0.71 ≈ 0.87 1

We now plot the above x and y values on the coordinate plane, as shown:
1.4 y 1.2 1.0 0.8 0.6 0.4 0.2

-0.2 -0.4 -0.6 -0.8 -1.0 -1.2 -1.4

0o

30o

60o

90o

120o 150o 180o 210o 240o 270o 300o 330o 360o

x

Unit Circle Trigonometry

Graphs of the Sine and Cosine Functions

Drawing a smooth curve through the points which we have plotted, we obtain the following graph of y = f ( x) = cos( x) :
1.4 y 1.2 1.0 0.8 0.6 0.4 0.2

y = f ( x) = cos( x)

-0.2 -0.4 -0.6 -0.8 -1.0 -1.2 -1.4

0o

30o

60o

90o

120o 150o 180o 210o 240o 270o 300o 330o 360o

x

Since coterminal angles occur every 360o, the cosine graph will repeat itself over and over again, as shown below: y 1

y = f ( x) = cos( x)

-360o -270o -180o

-90o

0o

90o

180o

270o

360o

450o

540o

630o

x 720o

-1

Unit Circle Trigonometry

Graphs of the Sine and Cosine Functions

The graphs can easily be used to determine the sine, cosine, secant, and cosecant of quadrantal angles.
Examples Use the graphs of the sine and cosine functions to find exact values of the following:

1. sin ( 90

)

Solution: If we look at the graph of y = sin( x) , where x = 90 , we find that the y-value is 1. Therefore, we conclude that sin 90 = 1 . 2. cos ( 270

( )

)

Solution: If we look at the graph of y = cos( x) , where x = 270 , we find that the yvalue is 0. Therefore, we conclude that cos ( 270 ) = 0 .

3. sec 180

(

)

Solution: If we look at the graph of y = cos( x) , where x = 180 , we find that the y-value is -1. This means that cos 180 = −1 . Since sec (θ ) = that sec (180 ) = 4. csc ( 360
1 cos 180

(

)

1 cos(θ )

, we conclude

(

)

=

1 −1

= −1 .

)

Solution: If we look at the graph of y = sin( x) , where x = 360 , we find that the y-value is 0. This means that sin(360 ) = 0 . Since csc (θ ) = sin1(θ ) , we conclude that csc ( 360

) is undefined, since

1 0

is undefined.

Exercises 1. Use the graphs of the sine and cosine functions to find exact values of the following: a) cos 270 b) sin −90 c) sec 90 d) csc 450

(

)

(

)

( )

(

)

2. Sketch the graphs of the following functions. Label all intercepts. a) y = f ( x) = sin( x) , where −90 ≤ x ≤ 270 b) y = f ( x) = cos( x) , where −720 ≤ x ≤ 90

Unit Circle Trigonometry

Graphs of the Sine and Cosine Functions