1. Section 3 – Circular Functions Objective To find the values of the six trigonometric functions of an angle in standard position given a point on the terminal side.

2. Sine and Cosine Functions Using the Unit Circle x y  P(x, y)= P(cosx, sinx) Consider an angle  in standard position. The terminal side of the angle intersects the circle at a unique point, P(x, y). The y-coordinate of this point is sine  The x-coordinate is cosine  The abbreviation for sine is sin. The abbreviation for cosine is cos.

3. Definition of Sin and Cos If the terminal side of an angle  in standard position intersects the unit circle at P(x, y), then cos  = x and sin  = y. Find sin 90° 90° Remember, the unit circle has a radius of 1. (0, 1) The terminal side of a 90° angle in standard position is the positive y-axis, what are the coordinates of the point where it intersects the unit circle? Since the sin  = y, what does sin 90° = 1

4. Sin and Cos Find cos   The terminal side of an angle of  radians is the negative x-axis. What are the coordinates of the point? (-1, 0) Since sin  = y. The cos  = 0

5. Sin and Cos When we use the unit circle, the radius, r, is 1. 1 P(x, y) Since sin  = y, we can say sin  y r Since cos  = x, we can say cos  = x r

6. Sin and Cos Functions of Any Angle in Standard Position For any in standard position with measure , a point P(x, y) on its terminal side, and r = √ x² + y², the sin and cos functions of  are as follows: For any in standard position with measure , a point P(x, y) on its terminal side, and r = √ x² + y², the sin and cos functions of  are as follows: sin  = y r cos  = x r

7. Find the values of the sin and cos functions of an angle in standard position with measure  if the point with coordinates (3, 4) lies on its terminal side. (3, 4) 3 4 r You know that x = 3 and y = 4. You need to find r. r = √ 3² + 4² r = √9 + 16 = 5 Now you know x = 3, y = 4, and r = 5. You can write the sin and cos functions. Sin = y/r, or 4/5. Cos = x/r, or 3/5.

8. Find sin  when cos  = 5/13 and the terminal side of  is in Quadrant 1. Since cos  = x/r = 5/13 and r is always positive, r = 13 and x = 5. You now have to find y. r = √x² + y² 13 = √5² + y² 169 = 25 + y² 144 = y² + 12 = y Since y is in Quadrant 1, y must be positive. So, sin  = y/r, or 12/13 13 5 y

9. There are 4 other trig functions – tangent (tan), cotangent (cot), secant (sec), and cosecant (csc). For any angle in standard position with measure , and a point P(x, y) on its terminal side the trig functions are: For any angle in standard position with measure , and a point P(x, y) on its terminal side the trig functions are: sin  = y/rcos  = x/rtan  = y/x cot  = x/ycsc  = r/ysec  = r/x

10. The terminal side of an angle in standard position contains the point with coordinates (8, -15). Find tan, cot, sec, and csc. Since x = 8 and y = -15, you must find r. Since x = 8 and y = -15, you must find r. r = √8² + (-15)² = √289 = √289 = 17 = 17 x = 8, y = -15, and r = 17. You can now write the trig functions. tan  = -15/8 cot  -8/15 sec  = 17/8 csc  = -17/15 8 -15 r

11. If csc  = -2 and  lies in Quadrant 3, find sin, cos, tan, cot, and sec. Since csc and sin are reciprocals, sin  = -½. To find the other functions, you have to find the coordinates of a point on the terminal side. Since sin  = -½ and r is always positive, let r = 2 and y = -1. Since sin  = -½ and r is always positive, let r = 2 and y = -1. Find x. r² = x² + y² 2² = x² + (-1)² 4 = x² + 1 3 = x² √3 = x Since the terminal side of  lies in quadrant 3, x = -√3 sin  = -√3/2 tan  = -1/-√3 sec  = 2/-√3 or 2√3/3 cot  = -√3/ -1

12. Assignment Page 260 – 261 Page 260 – 261 #22 – 41, 49, 50 #22 – 41, 49, 50