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Why does cos θ, which normally equals x, become 0 when you evaluate this function

  1. Jul 11, 2010 #1
    Why does cos θ, which normally equals x, become 0 when you evaluate this function for π/r radians?
    (π/r means pi over r radians)
    Why does cos θ, which normally equals x, become 0 when you evaluate this function for π/r radians?
    It says that the point on the circle for this question is (0,1) so zero = x and y = 1 here.
     
  2. jcsd
  3. Jul 11, 2010 #2

    Mentallic

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    For [tex]x=cos\theta[/tex], when x=0, [tex]\theta=\pi/2[/tex] (it can also be [tex]-\pi/2,3\pi/2[/tex] etc. but I don't think you need to deal with that right now).

    When you're looking at the circle and any point on the circle (x,y) remember that [tex]cos\theta[/tex] is adjacent/hypotenuse or x/1=x. So since the point on the circle is (0,1), the x value is 0 so [tex]cos(\pi/2)=0[/tex]
     
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