# HW12sol - MAC1114 - Homework 12 Due Thursday - April 7 1....

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Unformatted text preview: MAC1114 - Homework 12 Due Thursday - April 7 1. Recall that cos(u + v ) = cos u cos v − sin u sin v . (a) Show that cos(2u) = 1 − 2 sin2 u. cos(2u) = cos(u + u) = cos u cos u − sin u sin u = cos2 u − sin2 u = (1 − sin2 u) − sin2 u = 1 − 2 sin2 u 1 − cos 2u (b) Use part (a) to show that sin2 u = 2 cos(2u) = 1 − 2 sin2 u ⇒ 2 sin2 u = 1 − cos(2u) ⇒ sin2 u = 1−cos(2u) 2 (c) Use part (b) to show that, for v in quadrant I, sin v = 2 1 − cos v 2 Take the identity in part (b) and substitute u = v to get sin2 v = 1−cos v ⇒ sin v = ± 1−cos v . 2 2 2 2 2 Since v is in Quadrant I, we use the positive part (actually this is technically incorrect, I should π π have said 0 ≤ v ≤ π . The angle v = 94 is in quadrant I for example, but v = 98 is in quadrant 2 2 v III, so sin 2 < 0 and we'd want the negative piece). 2. Solve the equation tan x − sin x = 0 in the interval [0, 2π ). 2 See your class notes. 3. If cos x = 1 5 and x is in Quadrant I, nd sin 4x. Draw a triangle or use the identity sin2 x +cos2 x = 1 to nd that sin x = 2 2 sin(2 · 2x) = 2 sin(2x) cos(2x) = 2(2 sin x cos x)(cos x − sin x) = 2(2 1 √ 24 5 . Now note that sin(4x) √ 24 1 1 )( 25 − 24 ) = 4 24(−23) . 55 25 625 √ = ...
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## This note was uploaded on 06/10/2011 for the course MAC 1114 taught by Professor Gentimis during the Spring '11 term at University of Florida.

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