SECTION 6.3 Properties of the Trigonometric Functions 421 Finding the Values of the Trigonometric Functions of θ when the Value of One Function Is Known and the Quadrant of θ Is Known Given the value of one trigonometric function and the quadrant in which θ lies, the exact value of each of the remaining five trigonometric functions can be found in either of two ways. Option 1 Using a Circle of Radius r Step 1 Draw a circle centered at the origin showing the location of the angle θ and the point ( ) = P x y , that corresponds to θ. The radius of the circle that contains ( ) = P x y , is = + r x y . 2 2 Step 2 Assign a value to two of the three variables x , y , r based on the value of the given trigonometric function and the location of P . Step 3 Use the fact that P lies on the circle + = x y r 2 2 2 to find the value of the missing variable. Step 4 Apply the theorem on page 407 to find the values of the remaining trigonometric functions. Option 2 Using Identities Use appropriate identities to find the value of each remaining trigonometric function. Figure 45 θ = tan 1 2 ; θ in quadrant III 3 3 P 5 (22, 21) 23 23 y x x2 1 y2 5 5 u Given the Value of One Trigonometric Function and the Sign of Another, Find the Values of the Remaining Ones Given that θ = tan 1 2 and θ < sin 0, find the exact value of each of the remaining five trigonometric functions of θ. Solution Option 1 Using a Circle EXAMPLE 6 Step 1 Since θ = > tan 1 2 0 and θ < sin 0, the point ( ) = P x y , that corresponds to θ lies in quadrant III. See Figure 45. Step 2 Since θ = = y x tan 1 2 and θ lies in quadrant III, let = − x 2 and = − y 1. Step 3 With =− =− x y 2 and 1, then ( ) ( ) = + = − + − = r x y 2 1 5, 2 2 2 2 so P lies on the circle + = x y 5. 2 2 Step 4 Apply the theorem on page 407 using = − = − = x y r 2, 1, and 5. θ θ θ θ θ = = − = − = = − = − = = − = − = = − = − = = − − = y r x r r y r x x y sin 1 5 5 5 cos 2 5 2 5 5 csc 5 1 5 sec 5 2 5 2 cot 2 1 2 Option 2 Using Identities Because the value of θ tan is known, use the Pythagorean identity that involves θ tan , that is, θ θ + = tan 1 sec . 2 2 Since θ = > tan 1 2 0 and θ < sin 0, then θ lies in quadrant III, where θ < sec 0. θ θ θ θ θ ( ) + = + = = + = = − tan 1 sec 1 2 1 sec sec 1 4 1 5 4 sec 5 2 2 2 2 2 2 Pythagorean identity tan 1 2 θ = Proceed to solve for θ sec . sec 0 θ < (continued)
RkJQdWJsaXNoZXIy NjM5ODQ=