1. 2.1 Six Trig Functions for Right Triangles
sin () = Opposite csc () = Hypotenuse [cosecant]
Hypotenuse Opposite
cos () = Adjacent sec () = Hypotenuse [secant]
Hypotenuse Adjacent
tan () = Opposite cot() = Adjacent [cotangent]
Adjacent Opposite
Note:
 is an Acute angle.
sin () = 12 csc() = 13
13 12
cos() = 5 sec() = 13
13 5
tan() = 12 / 5 cot() = 5 / 12  13 12  5

2. Trig Cofunction Identities
sin () = cos (90 - ) cos () = sin (90 - )
tan () = cot (90 - ) cot () = tan (90 - )
sec () = csc (90 - ) csc () = sec (90 - )
90- c a  b

3. 2.2 Reference Angles
Evaluating trigonometric functions of angles greater than 90° as well as
negative angles is done by using a reference angle.
Let  be a nonacute angle in standard position that lies in a quadrant.
Its reference angle is the positive acute angle ’ formed by the terminal
side of  and the x-axis. 0°
90
180°
270° 0°
90
180°
270° 0°
90
180°
270°
Angle: 120°
Reference Angle: 60°
Angle: 320° or -40°
Reference Angle: 40°
Angle: 200° or -160°
Reference Angle: 20°
Find exact values for
sin(120) =
cos (120) =
tan (120) =

4. 2.3 Finding Trig Functions with Calculators
Make sure the calculator is in ‘DEGREE’ mode
(Ti-83/84 – Use the ‘Mode’ Button)
Examples:
Find sin (49º 12’)
12/60 = .2 so, find sin (49.2) =
2. Find sec (97.977º)
Secant is the reciprocal of cosine.
Calculators do not have a ‘secant’ function so
sec (97.977) = 1/cos(97.977) =
3. Sin  = , Find 
Sin-1 ( ) = 74.4º

5. 2.4 Solving Right TrianglesB 25 x 50 A C y
Find the length of side BC
Sin (50) = x
So, x = 25 Sin (50) = 25 (.766) = 19.15 25
How would you find side AC?
Think about using the sine, cosine or tangent function.

6. Angles of Elevation/Depression
Ground 
20,000 ft 
34,000 ft
Angle  is an angle of elevation from the ground to the sky.
Angle  is an angle of depression from the sky to the ground.
You can use trigonometric functions (sine, cosine, tangent) to find the angles & sides.
Tan () = 20,000 / 34,  = tan (20,000/34,000) = 30 -1

7. 2.5 Bearing Application Problems
Bearings are expressed using one of two methods:
Method 1: (A single angle is given in degrees)  220º
-- Measure the angle in a ‘clockwise’ direction from North
Method 2: (Directions and angle are given)  S 40º W
-- Start with North/South Line
-- Use an acute angle to show direction
either east or west from the line.
220º
40º

8. Example: Bearing Problem
The bearing from A to C is S 52º E.
The bearing from A to B is N 84º E
The bearing from B to C is S 38º W.
A plane flying at 250 mph takes 2.4 hours to go from A to B.
Find the distance from A to C.
Finding distance from A to B
D= RT
D= (250)(2.4) =600 miles
Finding distance from A to C
Sin (46) = AC
600
AC = 600Sin(46)
= 600 (.7193)
= miles
96º A
84º
600 miles B
46º
44º
38º
52º C