Section 4.2, Trigonometric Functions: The Unit Circle
Homework: 4.2 #1–41 odds
1 Trigonometric Functions
Instead of focusing on the angle, we will spend much of the semester focusing on the point ( x, y ) where the ray created by the angle crosses the unit circle.
First, note that x
2
+ y
2
= 1 by the Pythagorean Theorem. (We’ll discuss this in more detail later.)
Let θ be a real number and let ( x, y ) be the point on the unit circle corresponding to θ . Then, the six trigonometric functions (sine, cosine, tangent, cosecant, secant, and cotangent) are sin cos
θ
θ
=
= y x tan θ = y x
= sin θ cos θ
, x = 0 cot θ =
1 csc θ =
1 y sec θ = x x 1
= y tan θ
=
=
=
1 sin θ
, y = 0
1 cos θ
, x = 0 cos θ
, y = 0 sin θ
Examples
Calculate all 6 trigonometric functions for each of the following angles.
1.
θ = 3 π/ 4
This corresponds to the point ( −
√
2
2
, sin cos tan
3
4
3
π
4
π
3 π
4
√
2
=
=
2
√
= −
√
2
2
−
2 / 2
√
2 / 2
= − 1
√
2
2
) on the unit circle, so sec csc
3
4
π
3
4
=
π cot
−
3
=
4
π
1
√
2 / 2
1
√
2 / 2
=
1
=
√
2
= −
√
2
− 1
= − 1
(Be sure to rationalize fractions!)
2.
θ = π/ 2
This angle corresponds to (0 , 1) on the unit circle, so
π sin
2
π cos tan
2
π
2
= 1
= 0
=
1
0
= undefined sec
π
2
=
π 1 csc = = 1
2 1
1
= undefined
0 cot
π
2
=
0
1
= 0
3.
θ = 7 π/ 6
1

This angle corresponds to ( −
√
3
2
, − 1
2
), so sin cos
7
6
π
7 π
6
1
= −
= −
2
√
3
2 tan
7 π
6
− 1 / 2
=
− 3 / 2
=
3
=
√
3
3 sec
7 π
6
= csc
−
1
√
3 / 2
7 π
6
=
= −
2
√
3
1
= − 2
− 1 / 2
= −
2
√
3
3 cot
7 π
6
=
√
3
2 Some Notes on Trigonometric Functions
Note that
− 1 ≤ sin θ ≤ 1
− 1 ≤ cos θ ≤ 1
1 ≤ csc θ
1 ≤ sec θ
There are no limitations of values for the values of tan θ and cot θ .
The quadrant of the angle can help to determine the sign of the trigonometric functions:
Quadrant II 6 Quadrant I sin cos tan
θ >
θ <
θ <
0
0
0 sin θ > 0 cos θ > 0 tan θ > 0
sin θ < 0 cos θ < 0 tan θ > 0 sin θ < 0 cos θ > 0 tan θ < 0
Quadrant III Quadrant IV
All trigonometric functions are 2 π -periodic. For example, sin θ = sin( θ ± 2 π ) = sin( θ ± 4 π ) = · · · cos θ = cos( θ ± 2 π ) = cos( θ ± 4 π ) = · · ·
This occurs because adding or subtracting 2 π to an angle gives you a coterminal angle, so it can be represented by the same point on the unit circle.
Example
Evaluate each of the following:
1. sin
11 π
4
= sin
3 π
4
=
√
2
2
2. tan( − 5 π
6
) = tan
7 π
6
=
√
3
3
All of the trigonometric functions are either even or odd. cos and sec are even, so cos( − θ ) = cos θ sec( − θ ) = sec θ
The rest of the functions are odd, so sin( − θ ) = − sin θ tan( − θ ) = − tan θ csc( cot(
−
−
θ
θ
) =
) =
−
− csc cot
θ
θ
2

Examples
1. Given that sin( − θ ) = − 3 / 4, find
(a) sin θ = 3 / 4
(b) csc θ = 4 / 3
2. Given that cos θ = 2 / 5, find
(a) cos( − θ ) = 2 / 5
(b) sec( − θ ) = 5 / 2
Note: There are many other properties of trigonometric functions, but we have all semester to cover them!
Another Note: The book has some good diagrams on page 295.
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