csc θsec θcot θ...Find the value of each trig function. G) cos = H) tan = I) sin = π 3 π 6 π 4 1 2 √3 3 √2 2 Find the value of each trig function. G) csc = H) sec = I) cot

12.4 NOTES Trigonometric Functions of Any Angle

1

BELLWORK #1: Calculate the following trig functions. What do you notice?

A) sin(45°) =

B) sin(135°) =

C) sin(225°) =

D) sin(315°) =

√22

√22√22

√22

-

-

BELLWORK #2: Calculate the following trig functions. What do you notice?

A) cos(60°) =

B) cos(120°) =

C) cos(240°) =

D) cos(300°) =

12

12

-

12

-

12

BELLWORK #3: What Greek letter is this?

θ"theta"

y

θ

(x, y)

r

LESSON 12. 4 - Trig Functions of Any Angle

• Let θ be an angle in standard position and let (x, y) be the point where the terminal side of θ intersects the circle x2 + y2 = r2. • The six trigonometric functions of θ can are defined as follows:

x

sin θ = csc θ =yr

ry y ≠ 0

cos θ = sec θ =xr

rx x ≠ 0

tan θ = cot θ =yx

xy y ≠ 0x ≠ 0 = -0.75

= -1.6667

= 1.25

= -1.3333

= -0.6

Let (‑3, 4) be a point on the coterminal side of angle θ in standard position. Calculate the six trigonometric function of angle θ.

sin θ =

cos θ =

tan θ =

csc θ =

sec θ =

cot θ =

y

θ

(‑3, 4)r

x

45

= 0.8

-354-3545-3-34

-3

45

42 + (-3)2 = r2

16 + 9 = r2

25 = r2

5 = r


2

= -0.3846

Let (12, ‑5) be a point on the coterminal side of angle θ in standard position. Calculate the six trigonometric function of angle θ.

y

θ

(12, ‑5)r

x

sin θ =

cos θ =

tan θ =

csc θ =

sec θ =

cot θ =

-513

122 + (-5)2 = r2

144 + 25 = r2

169 = r2

13 = r

13-5

12 = 0.92311213

= -0.4167-512

= -2.613-5

= 1.08331312

= -2.412-5

Let (‑8, ‑15) be a point on the coterminal side of angle θ in standard position. Calculate the six trigonometric function of angle θ.

y

θ

(‑8, ‑15)r

x

= -0.8824sin θ =

cos θ =

tan θ =

csc θ =

sec θ =

cot θ =

-1517

(-8)2 + (-15)2 = r2

64 + 225 = r2

289 = r2

17 = r

= 0.4706-817

= 1.875-15-8

= -1.133317

-15

= -2.12517-8

= 0.5333-8

-15

-8

-15= 17

THE UNIT CIRCLE

• The circle x2 + y2 = 1, which has a center at (0, 0) and a radius length of 1, is called the UNIT CIRCLE.

• To simplify calculations, we like to place an angleʹs triangle into the unit circle, because its radius is 1. This means that the trig functions will only depend on what x and y are.

y

θ

(x, y)

r = 1

x

• In the unit circle, sin(θ) = y, cos(θ) = x, and tan(θ) = xy

COMMON ANGLES IN TRIGONOMETRY

• The angles 30°, 45°, and 60° occur frequently in trig. Those angles create special right triangles, whose sides are always in a certain ratio:

30°

60°

45°

45°

1

√3

2 √2 1

1Reduce each hypotenuse to be equal to 1 (and adjust other side

lengths proportionally) to make the Unit Circle ratios below:

Find the value of each trig function.

A) sin(45°) =

B) cos(60°) =

C) tan(30°) =

2212

√33


A) cot(45°) =

B) csc(60°) =

C) sec(30°) =

1

2√32

2√32


3


G) cos =

H) tan =

I) sin =

π3π6

π4

12

√33

√22


G) csc =

H) sec =

I) cot =

π3π6

π4

2√32

2√32

1

REFERENCE ANGLES

• Let θ be an angle in standard position. The

REFERENCE ANGLE for θ is the acute angle θʹ formed by the terminal side of θ and the x‑axis.

• Reference angles are only calculated for angles between 90° and 360°.

• If an angle is less than 90°, then it is already acute.

• If an angle is greater than 360° or less than or equal to 0°, first find its coterminal angle between 90° and 360°, then calculate the reference angle measure.

What is the reference angle?

θ = 100°θθ'

θʹ = 80°


θ = 205°

'θʹ = 25°


θ = 660°

'

θʹ = 60°


4


θ = ‑315°

'θʹ = 45°


θ = ‑570°

'θʹ = 30°


2π3θ =

'

θʹ = π3


7π4

θ =

'θʹ = π4


19π6θ =

'θʹ = π6


7π6

‑θ =

'

θʹ = π6


5


14π3

‑θ =

'θʹ = π3

REFERENCE ANGLES

• Reference angles allow you to find a trigonometric function for any angle θ.

• The sign of the trigonometric function will depend on in which quadrant θ lies.

Evaluate the trigonometric function using reference angles.

sin(495°)495° is coterminal with 135°, which is in Quadrant II.

The reference angle for 135° is 45°.

So, sin(45°) =

The sine function is positive in Quadrant II, so the final answer is

sin(495°) =

√22

√22


tan(570°)570° is coterminal with 210°, which is in Quadrant III.


So, tan(30°) =

The tangent function is positive in Quadrant III, so the final answer is

tan(570°) =

√33

√33


cos(‑240°)-240° is coterminal with 120°, which is in Quadrant II.


So, cos(60°) =

The cosine function is negative in Quadrant II, so the final answer is

cos(-240°) = -

12

12


sin(‑390°)-390° is coterminal with 330°, which is in Quadrant IV.


So, sin(30°) =

The sine function is negative in Quadrant IV, so the final answer is

sin(-390°) = -

12

12


6


8π3cos

is coterminal with , which is in Quadrant II.

The reference angle for is .

So, cos( ) =

The cosine function is negative in Quadrant II, so the final answer is

cos( ) = -

12

12

8π3

2π3

2π3

π3

π3

8π3


15π4sin

is coterminal with , which is in Quadrant IV.


So, sin( ) =

The sine function is negative in Quadrant IV, so the final answer is

sin( ) = -

15π4

7π4

7π4

π4

π4

22

√22

15π4


‑5π6tan

is coterminal with , which is in Quadrant III.


So, tan( ) =

The tangent function is positive in Quadrant III, so the final answer is

tan( ) =

-5π6

7π6

7π6

π6

π6

√33

-5π6

√33


‑11π3

cos

is coterminal with , which is in Quadrant I.

Since is already in Quadrant I, there is not need to find a reference angle.

So, cos( ) =

The cosine function is positive in Quadrant I, so the final answer is

cos( ) =

12

12

-11π3

π3

π3

π3

-11π3

HOMEWORK:12.4 Worksheet ‑ Trig Functions of Any Angle

Documents

csc θsec θcot θ...Find the value of each trig function. G) cos = H) tan = I) sin = π 3 π 6 π 4 1 2 √3 3 √2 2 Find the value of each trig function. G) csc = H) sec = I) cot