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Concavity of Trig Functions

  1. Jun 3, 2008 #1
    [SOLVED] Concavity of Trig Functions

    I'm working out a problem, but I'm stuck at a spot.

    The original equation is y = 2 cos x + sin 2x, and yes I did the second derivative, which came out to be

    cos x (4 sin x + 1)

    when i equate it to zero, i get
    cos x = 0 and 4 sin x + 1 = 0

    (Let's just say the domain was defined from 0 to 2 pi)

    For SOME reason, I keep trying to evaluate x when 4 sin x + 1 = 0

    by using inverse sin ....so x = inverse sin (-0.25)

    My calculator is in radians, but it still gives me an answer of -0.25. I used a graphing calculator, and the answer SHOULD be 3.39. What am I doing wrong?
     
  2. jcsd
  3. Jun 3, 2008 #2

    Defennder

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    Homework Helper

    You forgot that the third quadrant anti-clockwise of the ASTC diagram is also negative. And that corresponds to pi + the basic angle. Basic angle is given by arcsin 1/4.
     
  4. Jun 4, 2008 #3
    In your second derivative, I believe you are missing a constant

    y'=2(cos(2x)-sin(x)), then y'' = -2(2sin(2x)+cos(x)) = -2cos(x)(4sin(x)+1)
    By double-angle formula sin(2A)=2sin(A)cos(A)
     
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