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Math Help - equation of the tangent line with trig

  1. #1
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    equation of the tangent line with trig

    Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x))

    y'=cos[(sinx)(cosx)]
    y'3π=cos[(sinx)(-1)]
    cos [(0)(-1)]
    cos(0)=1


    y-0=1(x-3π)
    y= x-3
    πWhat's wrong with my answer?
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    Quote Originally Posted by hazecraze View Post
    Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x))

    y'=cos[(sinx)(cosx)]
    y'3π=cos[(sinx)(-1)]
    cos [(0)(-1)]
    cos(0)=1

    y-0=1(x-3π)
    y= x-3πWhat's wrong with my answer?

    Your derivative's wrong: (\sin({\sin x}))'=\cos({\sin x})\cos x.

    Tonio
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    Quote Originally Posted by hazecraze View Post
    Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x))

    y'=cos[(sinx)(cosx)]

    You wrote that wierd, so I'm not sure if you got it wrong, or just used bad notation.

    y'3π=cos[(sinx)(-1)]
    cos [(0)(-1)]
    cos(0)=1

    y-0=1(x-3π)
    y= x-3πWhat's wrong with my answer?
    The derivative is cos(sin(x))cos(x).

    y'(3\pi)=cos(sin(3\pi))cos(3\pi)=cos(0)(-1)=-1
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  4. #4
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    Quote Originally Posted by hazecraze View Post
    Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x))

    y'=cos[(sinx)(cosx)] <<<<<<<<< here. This line should read: y' = cos(sin(x)) * cos(x)
    y'3π=cos[(sinx)(-1)]
    cos [(0)(-1)]
    cos(0)=1


    y-0=1(x-3π)
    y= x-3
    πWhat's wrong with my answer?
    The equation of the tangent line becomes now: y = -x + 3\pi
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