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Math Help - equation of tangent

  1. #1
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    equation of tangent

    Can someone please check my answers below for accurateness?
    let me know if you get the same or different answer with me.

    1. Find an equation of the tangent line to the curve y=sinx+cos2x at the point (pi/6, 1)

    My final answer is:
    y=[-(sqrt3)x]/2 + 1 + [(sqrt3) pi]/12

    What did you get?

    2. Find the slope of the tangent line to the curve y=cos(cosx) at x=pi/2

    my final answer is y=1

    did you get this too?

    Thanks for your help!
    Last edited by skeske1234; August 18th 2009 at 11:12 AM.
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  2. #2
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    1. Are you sure about the function since y\left(\frac{\pi}{2}\right) \neq 1 ?

    2. yes but no ! This is the equation of the tangent whereas you are asked the slope
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  3. #3
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    I don't get the same answer for the first one.

    y=\sin(x)+\cos(2x)

    y'=\cos(x)-2\sin(2x)

    y-1=[\cos(\frac{\pi}{2})-2sin(\pi)](x-\frac{\pi}{2})

    There shouldn't be a slope. It is just a horizontal line. y=1. The point (\frac{\pi}{2},1) is an extreme point.
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  4. #4
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    Quote Originally Posted by adkinsjr View Post
    I don't get the same answer for the first one.

    y=\sin(x)+\cos(2x)

    y'=\cos(x)-2\sin(2x)

    y-1=[\cos(\frac{\pi}{2})-2sin(\pi)](x-\frac{\pi}{2})

    There shouldn't be a slope. It is just a horizontal line. y=1. The point (\frac{\pi}{2},1) is an extreme point.
    and SO SORRY! typed question wrong in number 1, it is bolded now if you scroll up to check.. The point was actually (pi/6, 1).
    so if this is the case, would my answer be correct now?

    here is my work:

    y' = cos (pi/6) - 2 sin (pi/3)
    =-sqrt3/2

    now y=mx+b I have to use this (says my teacher)
    1=(-sqrt3/2) (pi/6) + b
    isolate for b and I get
    1+ (sqrt3/12)(pi) = b

    therefore eqtn is
    y=-sqrt3/2 x + 1+ (sqrt3/12) pi
    Last edited by skeske1234; August 18th 2009 at 11:14 AM.
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