This document discusses trigonometric functions of angles. It defines reference angles for quadrants I through IV and provides examples of finding reference angles for given angles. It also reviews Pythagorean identities for trigonometric functions and shows how to use trigonometric functions like sine to find the area of a triangle when given angle measures and side lengths. Finally, it provides an example of finding the exact value of secant 120 degrees without a calculator.

1. 6.3 Trigonometric Functions of Angles

2. 6.3 Trigonometric Functions of Angles
Much of this is review and is assumed.

3. 6.3 Trigonometric Functions of Angles
Much of this is review and is assumed.
Galatians 5:22-23 But the fruit of the Spirit is love, joy,
peace, patience, kindness, goodness, faithfulness, gentleness
and self-control. Against such things there is no law.

4. Reference Angles (using degrees)

5. Reference Angles (using degrees)
Q I rotation
θ'=θ

6. Reference Angles (using degrees)
Q I rotation Q II rotation
θ'=θ θ ' = 180° − θ

7. Reference Angles (using degrees)
Q III rotation
θ ' = θ − 180°

8. Reference Angles (using degrees)
Q III rotation Q IV rotation
θ ' = θ − 180° θ ' = 360° − θ

9. Find the reference angles for each of the following:
1) 327°
2) 57°
3) 107°
4) 227°

10. Find the reference angles for each of the following:
1) 327° 33° QIV 360 − 327
2) 57° 57° QI
3) 107° 73° QII 180 − 107
4) 227° 47° QIII 227 − 180

11. Find the reference angles for each of the following:
1) 327° 33° QIV 360 − 327
2) 57° 57° QI
3) 107° 73° QII 180 − 107
4) 227° 47° QIII 227 − 180
Questions???

12. Pythagorean Identities revisited

13. Pythagorean Identities revisited
2 2
sin θ + cos θ = 1
2 2
sin θ = 1− cos θ
2 2
cos θ = 1− sin θ

14. Pythagorean Identities revisited
2 2
sin θ + cos θ = 1
2 2
sin θ = 1− cos θ
2 2
cos θ = 1− sin θ
2 2
tan θ + 1 = sec θ
2 2
tan θ = sec θ − 1
2 2
1 = sec θ − tan θ

15. Pythagorean Identities revisited
2 2
sin θ + cos θ = 1
2 2
sin θ = 1− cos θ
2 2
cos θ = 1− sin θ
2 2
tan θ + 1 = sec θ
2 2
tan θ = sec θ − 1
2 2
1 = sec θ − tan θ
2 2
1+ cot θ = csc θ
2 2
cot θ = csc θ − 1
2 2
1 = csc θ − cot θ

16. Pythagorean Identities revisited
2 2
sin θ + cos θ = 1
2
sin θ = 1− cos θ 2 Learn to recognize all
2 2 9 forms of the
cos θ = 1− sin θ
2 2
Pythagorean Identities
tan θ + 1 = sec θ
2 2
tan θ = sec θ − 1
2 2
1 = sec θ − tan θ
2 2
1+ cot θ = csc θ
2 2
cot θ = csc θ − 1
2 2
1 = csc θ − cot θ

17. Consider this scalene triangle:

18. Consider this scalene triangle:
1
A = bh
2

19. Consider this scalene triangle:
1
A = bh
2
h
and sin θ =
a

20. Consider this scalene triangle:
1
A = bh
2
h
and sin θ =
a
∴ h = asin θ

21. Consider this scalene triangle:
1
A = bh
2
h
and sin θ =
a
∴ h = asin θ
1
A = absin θ
2

22. Consider this scalene triangle:
1
A = bh
2
h
and sin θ =
a
∴ h = asin θ
1
A = absin θ
2
θ is the included angle between a & b (SAS relationship)
and other variations are also possible ...

23. Find the area of this triangle:
3
120°
18

24. Find the area of this triangle:
3
120°
18
1
A = absin θ
2

25. Find the area of this triangle:
3
120°
18
1
A = absin θ
2
1
A = ( 3) (18 ) sin120°
2

26. Find the area of this triangle:
3
120°
18
1
A = absin θ
2
1
A = ( 3) (18 ) sin120°
2
A ≈ 23.4

27. Find the exact value of: sec 120° (no calculator)

28. Find the exact value of: sec 120° (no calculator)
The reference angle is 60° in QII.
2π
This is on the Unit Circle.
3
2π 1
Cos at is − so ...
3 2
sec 120° = − 2

29. HW #5
Quiz Tomorrow!!
Confidence comes not from always being right but from
not fearing to be wrong.
Peter T. Mcintyre