4.4 Trig Functions of Any Angle Reference Angles Trig functions in any quadrant

Angles Beyond 90°  Expand from the context of angles of a right triangle  Consider a ray from the origin through a given point (x, y) (x, y) r θ

Reference Triangle  Drop a perpendicular from (x,y) to the x-axis This forms a reference triangle (x, y) θ β

Definition of a Reference Angle Let  be a nonacute angle in standard position that lies in a quadrant. Its reference angle is the positive acute angle ´ (prime) formed by the terminal side or  and the x-axis. In quadrant II: θ’=180°-θ In quadrant III: θ’=θ-180° In quadrant IV: θ’=360°-θ

Angles in Different Quadrants  Note that x and y have different signs in the various quadrants Thus the trig functions will have different signs for the quadrants  In quadrant III, β=θ-∏ (-x, -y) θ β

Example a b   a b P(a,b) Find the reference angle , for the following angle:  =315º Solution:  =360 º º = 45 º

Angles in Different Quadrants  Remember "allsintancos" I II III IV all positive sin is positive Cos is negative Tan is negative tan is positive Sin is negative Cos is negative cos is positive Sin is negative Tan is negative

Standard Angles  Recall table of functions of basic angles  Now we can expand this table

Standard Angles sin cos tan 120° 135° 150° 180°

Let P = (-3, -4) be a point on the terminal side of . Find each of the six trigonometric functions of . Solution. We need values for x, y, and r to evaluate all six trigonometric functions. We are given the values of x and y. Because P = (-3, -4) is a point on the terminal side of , x = -3 and y = -4.We can use pythagorean theorem to find r, r = 5 x = -3y = -4 P = (-3, -4)  x y Example

The bottom row shows the reciprocals of the row above. Text Example Cont. Solution Now that we know x, y, and r, we can find the six trigonometric functions of .

Use reference angles to find the exact value of Solution a. We use our two-step procedure to find sin 135°. Step 1 Find the reference angle,  ´, and sin  ´. 135 º terminates in quadrant II with a reference angle  ´ = 180 º – 135 º = 45 º. x y 135° 45° sin 135° Example