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

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
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. On your birthday last year, the high temperature was 12 °C, and the low of temperature of 2 °C occurred at 4 am. What is the value of t when the temperature is at 12 °C? A. 10 = t B. 12 = t C. 14 = t D. 16 = t E. 18 = t
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern