SECTION 6.2 Trigonometric Functions: Unit Circle Approach 401 There is no need to memorize Table 2.To find the value of a trigonometric function of a quadrantal angle, draw the angle and apply the definition, as we did in Example 2. Finding Exact Values of the Trigonometric Functions of Quadrantal Angles Find the exact value of each of the following angles: (a) sin 3π ( ) (b) cos 270 ( ) − ° EXAMPLE 3 Solution (a) See Figure 24(a). The point P on the unit circle that corresponds to 3 θ π = is P 1, 0 , ( ) = − so y sin 3 0. π ( ) = = (b) See Figure 24(b). The point P on the unit circle that corresponds to 270 θ = − ° is P 0, 1 , ( ) = so x cos 270 0. ( ) − ° = = Figure 24 3 θ π = 1 1 y x P 5 (21, 0) 21 21 u 5 3p (a) 270 θ =− ° 1 21 21 1 y x P 5 (0, 1) u 5 2270° (b) Now Work PROBLEMS 21 AND 61 3 Find the Exact Values of the Trigonometric Functions of π = ° 4 45 Finding the Exact Values of the Trigonometric Functions of π = ° 4 45 Find the exact values of the six trigonometric functions of 4 45 . π = ° Solution EXAMPLE 4 We seek the coordinates of the point P x y , ( ) = on the unit circle that corresponds to 4 45 . θ π = = ° See Figure 25. First, observe that P lies on the line y x. = (Do you see why? Since P 45 1 2 90 , θ = ° = ⋅ ° must lie on the line that bisects quadrant I.) Since P x y , ( ) = also lies on the unit circle, x y 1, 2 2 + = it follows that + = + = = = = = x y x x x x y 1 1 2 1 1 2 2 2 2 2 2 2 2 2 2 Then Figure 25 4 45 θ π = = ° 1 21 21 1 458 y x P 5 (x, y) y 5 x x2 1 y2 5 1 y x x y , 0, 0 = > > sin 4 sin45 2 2 cos 4 cos45 2 2 tan 4 tan45 2 2 2 2 1 csc 4 csc45 1 2 2 2 sec 4 sec45 1 2 2 2 cot 4 cot45 2 2 2 2 1 π π π π π π = ° = = ° = = ° = = = ° = = = ° = = = ° = =
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