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

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

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) =

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) = .756995 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) = -7.205879 3. Sin  = .967709, Find  Sin-1 (.967709) = 74.4º

2.4 Solving Right Triangles B 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.

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 / 34,000  = tan (20,000/34,000) = 30 -1

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

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) = 431.58 miles 96º A 84º 600 miles B 46º 44º 38º 52º C