1

1

2 12

3

*Special Angles

45°

45°

60°

30°

30°, 45°, and 60° → common reference anglesMemorize their trigonometric functions.

Use the Pythagorean Theorem; triangles below.

12

3

*Special Angles

60°

30°

30°, 45°, and 60° → common reference anglesMemorize their trigonometric functions.

Use the Pythagorean Theorem; triangles below.

1

1

2

*Special Angles

45°

45°

30°, 45°, and 60° → common reference anglesMemorize their trigonometric functions.

Use the Pythagorean Theorem; triangles below.

*Special Angles

θ 0º 30º 45º 60º 90º

sin θ

cos θ

tan θ

RECIPROCAL TRIG FUNCTIONS

• SIN CSC

• COS SEC

• TAN COT

RECIPROCAL TRIG FUNCTIONS

• SIN CSC

• COS SEC

• TAN COT

Find trig functions of 300° without calculator.

Reference angle is 60°[360° 300°]; IV quadrant

300°

60°

sin 300° = cos 300° =tan 300° =csc 300° =sec 300° =cot 300° =

Use special angle chart.

Find trig functions of 120° without calculator.

Reference angle is 120°[180° 120°]; IIquadrant

sin 120° = cos 120° =tan 120° =csc 120° =sec 120° =cot 120° =

Use special angle chart.

60°

Find trig functions of 210° without calculator.

Reference angle is [210°-180°]; III quadrant

60°

sin 210° = cos 210° =tan 210° =csc 210° =sec 210° =cot 210° =

Use special angle chart.

sin 300° =

cos 300° =

tan 300° =

csc 300° =

sec 300° =

cot 300° =

12*Quadrant Angles

•Reference angles cannot be drawn forquadrant angles 0°, 90°, 180°, and 270°

•Determined from the unit circle; r = 1

•Coordinates of points (x, y) correspond to (cos θ, sin θ)

0° (1,0)180° (1,0)

*Quadrant Angles

90° (0,1) → (cos θ, sin θ)

270° (0,1)

r = 1

*Quadrant Angles

θ 0º 90º 180º 270º

sin θ

cos θ

tan θ 0 0

10

1

0

sinθcosθ

Find trig functions for 90°.

Reference angle is (360° 90°) → 270°

sin 270° =cos 270° =tan 270°=csc 270° =sec 270°=cot 270° =

270°

90°

Use quadrant angle chart.

*Coterminal Angle

θ1 = 405º

The angle between 0º and 360º having the same terminal side as a given angle.Ex. 405º 360º = coterminal angle 45º

θ2 = 45º

*Coterminal Angles

•Example cos 900° = (See quadrant angles chart)

Used with angles greater than 360°, or angles less than 0°.

•Example tan (135° ) =

(See special angles chart)

Find the value of sec 7π / 4

SOLUTION

Express as a function of the reference angle and find the value.

tan 210° sec 120 °

SOLUTION

Express as a function of the reference angle and find the value.

csc 225°sin (330°)

SOLUTION

Express as a function of the reference angle and find the value.

cos (5π) cot (9π/2)

SOLUTION

SIN COS TAN SEC CSC COT

0

30

45

60

90

Inverse Trig Functions

Used to find the angle•when two sides of right triangle are known...•or if its trigonometric functions are known

Notation used: 1sin x or arcsin x-

1tan x or arctan x-1cos x or arccos x-

Read: “angle whose sine is …”

Also,

Inverse trig functions have only one principal value of the angle

defined for each value of x:

90° < arcsin < 90° 0° < arccos < 180°

90° < arctan < 90°

Example:Given tan θ = 1.600, find θ to the nearest 0.1° for 0° < θ < 360°

•Tan is negative in II & IV quadrants1 1

1

tan tan = tan 1.600

tan 1.600

reference angle

- -

-

θ = 180° 58.0° = 122° II

θ = 360° 58.0° = 302° IV

Note: On the calculator entering 1tan 1.600 - results in 58.0°

Given sin θ = , find θ to the nearest 0.1° for 0° < θ < 360°

SOLUTION

Given cos θ = 0.0157, find θ to the nearest 0.1° for 0° < θ < 360°

SOLUTION

Given sec θ = 1.553 where sin θ < 0, find θ to the nearest 0.1° for 0° < θ < 360°

SOLUTION

Given the terminal side of θ passes through point (2, -1), find θ the nearest tenth for 0° < θ < 360°

SOLUTION

Given the terminal side of θ passes through point (3, 5), find θ the nearest tenth for 0° < θ < 360°

SOLUTION

Given the terminal side of θ passes through point (8, 8), find θ the nearest tenth for 0° < θ < 360°

SOLUTION

Given the terminal side of θ passes through point (-5, 12), find θ the nearest tenth for 0° < θ < 360°

SOLUTION

The voltage of ordinary house current is expressed as V = 170 sin 2πft , where f = frequency = 60 Hz and t = time in seconds.

• Find the angle 2πft in radians when V = 120 volts and 0 < 2πft < 2π

SOLUTION

• Find t when V = 120 volts

SOLUTION

The angle β of a laser beam is expressed as:

where w = width of the beam (the diameter)and d = distance from the source.Find β if w = 1.00m and d = 1000m.

12 tan2

wβ =

d-

SOLUTION