Stitz-Zeager_College_Algebra_e-book

We put theorem 107 to good use in the following

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: er-clockwise (π/6) = 12 of a revolution. Thus α is a Quadrant I angle. Coterminal 2π angles θ are of the form θ = α + 2π · k , for some integer k . To make the arithmetic a bit easier, we note that 2π = 12π , thus when k = 1, we get θ = π + 12π = 13π . Substituting 6 6 6 6 k = −1 gives θ = π − 12π = − 11π and when we let k = 2, we get θ = π + 24π = 25π . 6 6 6 6 6 6 π/ π 2. Since β = − 43 is negative, we start at the positive x-axis and rotate clockwise (42π3) = 2 of 3 a revolution. We find β to be a Quadrant II angle. To find coterminal angles, we proceed as π π π π before using 2π = 63 , and compute θ = − 43 + 63 · k for integer values of k . We obtain 23 , 10π 8π − 3 and 3 as coterminal angles. y y 4 4 3 3 2 2 1 −4 −3 −2 −1 −1 α= 1 2 3 4 −2 −3 −4 α= π 6 1 π 6 in standard position. x −4 −3 −2 −1 −1 π β = − 43 1 2 3 4 x −2 −3 −4 π β = − 43 in standard position. π 3. Since γ = 94 is positive, we rota...
View Full Document

Ask a homework question - tutors are online