154E2-F2011

154E2-F2011 - MA 15400 Fall 2011 Exam 2 sin u v = sin u cos...

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Unformatted text preview: MA 15400 Fall 2011 Exam 2 sin( u + v ) = sin u cos v + cos u sin v sin( u − v ) = sin u cos v − cos u sin v cos( u + v ) = cos u cos v − sin u sin v cos( u − v ) = cos u cos v + sin u sin v tan( u + v ) = tan u + tan v 1 − tan u tan v tan( u − v ) = tan u − tan v 1 + tan u tan v sin(2 u ) = 2sin u cos u cos(2 u ) = cos 2 u − sin 2 u tan(2 u ) = 2tan u 1 − tan 2 u sin 2 θ + cos 2 θ = 1 1 + tan 2 θ = sec 2 θ 1 + cot 2 θ = csc 2 θ A B C c a b α β γ MA 15400 Exam 2 Fall 2011 Covers Lessons 10-21, all of Sections 6.5, 6.7, 7.2, 7.3, 7.4, and 7.6 up to question #14 1 1. Which graph best represents the function ? 2 sin 3 − = π π x y 2. Given right triangle ABC, with ° = 90 γ , express side b in terms of angle α and side a . A. α tan a b = B. α csc a b = C. α cos a b = D. α sin a b = E. α cot a b = A B C D MA 15400 Exam 2 Fall 2011 Covers Lessons 10-21, all of Sections 6.5, 6.7, 7.2, 7.3, 7.4, and 7.6 up to question #14 2 3. Scientists sometimes use the formula ( ) ( ) d c bt a t f + + = sin to simulate temperature variations during the days, with time t in hours, temperature f(t) in °C, and t = 0 corresponding to midnight. Assume that f(t) is decreasing at midnight....
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154E2-F2011 - MA 15400 Fall 2011 Exam 2 sin u v = sin u cos...

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