Stitz-Zeager_College_Algebra_e-book

# This is represented schematically below horizontal

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online