Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: ed 6 using a triangle approach. Letting P (x, y ) denote the point on the terminal side of θ which lies on the Unit Circle, we drop a perpendicular line segment from P to the x-axis to form a √ 30◦ − 60◦ − 90◦ right triangle. After a bit of Geometry3 we find x = y 3. Since P (x, y ) lies √ on the Unit Circle, we substitute x = y 3 into x2 + y 2 = 1 to get 4y 2 = 1, or y = ± 1 . Here, 2 √ y > 0, so y = sin π = 1 , and since x = y 3, x = cos π = 23 . 6 2 6 y 1 P (x, y ) P (x, y ) θ= 60◦ π 6 y x 1 θ= π 6 = 30◦ x 5. Plotting θ = 60◦ in standard position, we find it is not a quadrantal angle and set about using a triangle approach. Once again, we get a 30◦ − 60◦ − 90◦√right triangle and, after the usual computations, find x = cos (60◦ ) = 1 and y = sin (60◦ ) = 23 . 2 y 1 P (x, y ) P (x, y ) 30◦ θ= y 60◦ x 1 θ = 60◦ x 3 Again, can you show this? 10.2 The Unit Circle: Cosine and Sine 615 In Example 10.2.1, it was quite easy to find the cosine a...
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