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Trigonometry in this calculus problem

  1. Oct 18, 2006 #1
    Find all the points on the graph of the function [tex]f(x) = 2 \sin x + \sin^2 x[/tex] at which the tangent is horizontal.
    [tex]f'(x) = 2 \cos x + 2 \sin x \cos x = 0[/tex]
    [tex]2 \cos x (1 + \sin x) = 0[/tex]
    [tex]\cos x = 0[/tex] or [tex]\sin x = -1[/tex]
    [tex]x = \frac{\pi}{2} + n\pi[/tex] or [tex]x = \frac{3\pi}{2} + 2n\pi[/tex]
    I then plug in to find the y-coordinates. However, the book's answer for the cos x = 0 part is [tex]x = \frac{\pi}{2} + 2n\pi[/tex]. But cos x = 0 for all [tex]x = \frac{\pi}{2} + n\pi[/tex]... right??? where did the 2 come from????
     
    Last edited: Oct 18, 2006
  2. jcsd
  3. Oct 18, 2006 #2
    [tex] f'(x) = 2\cos x + 4\sin x \cos x [/tex]

    [tex] 2\cos x(1+2\sin x) = 0 [/tex]
     
    Last edited: Oct 18, 2006
  4. Oct 18, 2006 #3
    oops. It's actually [tex]f(x) = 2 \sin x + \sin^2 x[/tex]
     
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