SECTION 6.2 Trigonometric Functions: Unit Circle Approach 405 Now Work PROBLEM 47 Solution (a) From Figure 30 on the previous page, we see the point 2 2 , 2 2 − − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ corresponds to 5 4 , π so π = = − x cos 5 4 2 2 . (b) Since 135 3 4 , π ° = the point 2 2 , 2 2 − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ corresponds to 135 ,° so sin135 2 2 . ° = (c) Since 315 7 4 , π ° = the point 2 2 , 2 2 − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ corresponds to 315 ,° so tan315 2 2 2 2 1. ° = − = − (d) The point − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ 2 2 , 2 2 corresponds to 4 , π − so sin 4 2 2 . π ( ) − = − (e) The point − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ 2 2 , 2 2 corresponds to 11 4 , π so cos 11 4 2 2 . π = − Finding Exact Values for Multiples of 6 30 π ° = or 3 60 π ° = Based on Figures 31 and 32, we see that (a) cos210 cos 7 6 3 2 π ° = = − (b) sin 60 sin 3 3 2 π ( ) ( ) − ° = − = − (c) tan 5 3 3 2 1 2 3 π = − = − (d) cos 8 3 cos 2 3 1 2 π π = = − EXAMPLE 10 CAUTION On your calculator the second functions sin , cos , 1 1 − − and tan 1− do not represent the reciprocal of sin, cos, and tan. j Now Work PROBLEMS 51 AND 55 The use of symmetry also provides information about certain integer multiples of the angles 6 30 π = ° and 3 60 . π = ° See Figures 31 and 32. Figure 31 21 21 3 –– 2 1 – 2 2 , ) ( 3 –– 2 1 – 2 ( , ) 1 – 2 3 –– 2 2 2 ) , ( 1 – 2 3 –– 2 2 ) ( , 7p––– 6 5p––– 6 11p––– 6 p–– 6 1 1 y x Figure 32 4p––– 3 1 21 21 1 y x 3 –– 2 2 2 2 2 ) , ( 3 –– 2 1 – 2( , ) 1 – 2 3 –– 2 ) ( , 1 – 2 3 –– 2 ) ( , 2p––– 3 5p––– 3 p–– 3 1 2 6 Use a Calculator to Approximate the Value of a Trigonometric Function Before getting started, you must first decide whether to enter the angle in the calculator using radians or degrees and then set the calculator to the correct MODE. Check your instruction manual to find out how your calculator handles degrees and radians. Your calculator has keys marked sin , cos , and tan . To find the values of the remaining three trigonometric functions, secant, cosecant, and cotangent, use the fact that, if P x y , ( ) = is a point on the unit circle on the terminal side of ,θ then x y x y y x sec 1 1 cos csc 1 1 sin cot 1 1 tan θ θ θ θ θ θ = = = = = = =
RkJQdWJsaXNoZXIy NjM5ODQ=