540 CHAPTER 5 Trigonometric Functions EXAMPLE 5 Using the Reciprocal Identities Find each function value. (a) cos u, given that sec u = 5 3 (b) sin u, given that csc u = - 212 2 SOLUTION (a) We use the fact that cos u is the reciprocal of sec u. cos u = 1 sec u = 1 5 3 = 1 , 5 3 = 1 # 3 5 = 3 5 Simplify the complex fraction. (b) sin u = 1 csc u sinu is the reciprocal of csc u. = 1 - 212 2 Substitute csc u = - 212 2 . = - 22 12 Simplify the complex fraction as in part (a). = - 2 223 212 = 24 # 3 = 223 = - 12 3 Divide out the common factor 2. = - 12 3 # 23 23 Rationalize the denominator. = - 23 3 Multiply. S Now Try Exercises 87 and 93. II Sine and cosecant positive x < 0, y > 0, r > 0 I All functions positive x > 0, y > 0, r > 0 III Tangent and cotangent positive x < 0, y < 0, r > 0 IV Cosine and secant positive x > 0, y < 0, r > 0 x y 0 Signs of Trigonometric Function Values U in Quadrant sinU cos U tanU cot U sec U csc U I + + + + + + II + − − − − + III − − + + − − IV − + − − + − Signs and Ranges of Function Values In the definitions of the trigonometric functions, r is the distance from the origin to the point 1x, y2. This distance is undirected, so r 70. If we choose a point 1x, y2 in quadrant I, then both x and y will be positive, and the values of all six functions will be positive. A point 1x, y2 in quadrant II satisfies x 60 and y 70. This makes the values of sine and cosecant positive for quadrant II angles, while the other four functions take on negative values. Similar results can be obtained for the other quadrants. This important information is summarized here. NOTE Because numbers that are reciprocals always have the same sign, the sign of a function value automatically determines the sign of the reciprocal function value.
RkJQdWJsaXNoZXIy NjM5ODQ=