1. Circular
Trigonometric
Functions

2. Special
Angles

3. *Special Angles 30°, 45°, and 60° → common reference angles
Memorize their trigonometric functions.
Use the Pythagorean Theorem; triangles below.
60°
45° 2 1 1
45°
30° 1

4. *Special Angles θ
30º
45º
60º
sin θ
cos θ
tan θ

5. *Special Angles θ 30º 45º 60º sin θ cos θ tan θ 0.7071 0.8660 0.5774
0.5000
0.7071
0.8660
cos θ
tan θ
0.5774
1.0000
1.7320

6. Find trig functions of 300° without calculator.
Reference angle is 60°[360° - 300°]; IV quadrant
sin 300° = - sin 60°
cos 300° = cos 60°
tan 300° = - tan 60°
csc 300° = - csc 60°
sec 300° = sec 60°
cot 300° = - cot 60°
60°
300°
Use special angle chart.

7. sin 300° = - sin 60° = =
cos 300° = cos 60° = = 1/ 2
tan 300° = - tan 60° = =
csc 300° = - csc 60° = =
sec 300° = sec 60° = = 2/1
cot 300° = - cot 60° = =

8. Quadrant
Angles

9. *Quadrant Angles Reference angles cannot be drawn for
quadrant angles 0°, 90°, 180°, and 270°
Determined from the unit circle; r = 1
Coordinates of points (x, y)
correspond to (cos θ, sin θ)

10. *Quadrant Angles 90° (0,1) → (cos θ, sin θ) 180° (-1,0) 0° (1,0)
270° (0,-1)

11. *Quadrant Angles θ 0º
90º
180º
270º
sin θ 1 -1
cos θ
tan θ

12. Find trig functions for - 90°.
Reference angle is (360° - 90°) → 270°
sin 270° = -1
cos 270° = 0
tan 270° undefined
csc 270° = -1
sec 270° undefined
cot 270° = 0
-90°
Use quadrant angle chart.
270°

13. Coterminal
Angles

14. *Coterminal Angle
The angle between 0º and 360º having the same terminal side as a given angle.
Ex. 405º - 360º =
coterminal angle 45º
θ1 = 405º
θ2 = 45º

15. *Coterminal Angles Used with angles greater than 360°,
or angles less than 0°.
Example
cos 900° =
cos (900° - 720°) =
cos 180° = -1
(See quadrant angles chart)

16. Example
tan (-135° ) =
tan (360° -135°) =
tan 225° =
LOOK→ tan (225° - 180°)
tan 45° = 1
(See special angles chart)

17. Convert from radian to degrees: sec [(7π/ 4)(180/ π)] = sec 315°
Find the value of sec 7π / 4
Convert from radian to degrees:
sec [(7π/ 4)(180/ π)] = sec 315°
SOLUTION

18. Angle in IV quadrant: sec →positive
sec (360° - 315°) = sec 45°
= 1 /(cos 45°) = √2 = 1.414
Look how this problem was worked
in previous lesson.
SOLUTION

19. Practice

20. Express as a function of the reference angle and find the value.
tan 210°
sec 120 °
SOLUTION

21. Express as a function of the reference angle and find the value.
sin (- 330°)
csc 225°
SOLUTION

22. Express as a function of the reference angle and find the value.
cos (-5π)
cot (9π/2)
SOLUTION

23. Inverse
Trigonometric
Functions

24. 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:
Read: “angle whose sine is …”
Also,

25. 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°

26. Example:
Given tan θ = , find θ
to the nearest 0.1° for 0° < θ < 360°
Tan is negative in II & IV quadrants

27. θ = 180° ° = 122° II
θ = 360° ° = 302° IV
Note: On the calculator entering
results in -58.0°

28. More
Practice

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

30. to the nearest 0.1° for 0° < θ < 360°
Given cos θ = , find θ
to the nearest 0.1° for 0° < θ < 360°
SOLUTION

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

32. Given the terminal side of θ passes
through point (2, -1), find θ the nearest 0.1° for 0° < θ < 360°
SOLUTION

33. Application

34. 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

35. Find t when V = 120 volts
SOLUTION

36. 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.
SOLUTION