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    Hi the question is:

    The curve with equation y=(sin 1/2x)/2+cos 1/2x is defined for 0<x<2pi.

    Show that dy/dx=(1/2 + cos 1/2x)/(2+cos 1/2x)^2

    I've used the quotient rule and ended up with
    dy/dx=(2+cos 1/2x((cos 1/2x)/2)-sin 1/2x((-sin 1/2x)/2))/(2+cos 1/2x)^2
    and not sure where to go from there. Would really appreciate some help please, thanks
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    (Original post by jordanwu)
    bump!
    Is the function  y = \frac{ \sin \frac{x}{2}}{2 + \cos \frac{x}{2}} ? Typed maths does not read very well :-)
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    (Original post by 16Characters....)
    Is the function  y = \frac{ \sin \frac{x}{2}}{2 + \cos \frac{x}{2}} ? Typed maths does not read very well :-)
    Yes it is
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    (Original post by jordanwu)
    Hi the question is:

    The curve with equation y=(sin 1/2x)/2+cos 1/2x is defined for 0<x<2pi.

    Show that dy/dx=(1/2 + cos 1/2x)/(2+cos 1/2x)^2

    I've used the quotient rule and ended up with
    dy/dx=(2+cos 1/2x((cos 1/2x)/2)-sin 1/2x((-sin 1/2x)/2))/(2+cos 1/2x)^2
    and not sure where to go from there. Would really appreciate some help please, thanks
    Write out the numerator again, then multiply out the sin and cos and apply two identities to the resulting terms.
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    (Original post by jordanwu)
    Yes it is
    Then yeah you have applied the quotient rule correctly and should do what SeanFM said.
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    (Original post by SeanFM)
    Write out the numerator again, then multiply out the sin and cos and apply two identities to the resulting terms.
    I'm not sure what cos 1/2x multiplied by cos 1/2x is...
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    Anyone?
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    (Original post by jordanwu)
    Anyone?
    cos squared of x/2 (sorry I don't know the latex for cos squared)
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    (Original post by jordanwu)
    I'm not sure what cos 1/2x multiplied by cos 1/2x is...
    (Original post by 16Characters....)
    Then yeah you have applied the quotient rule correctly and should do what SeanFM said.
    Woops! 16Characters was kindly helping you already so I shall leave this one to him.
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    (Original post by jordanwu)
    I'm not sure what cos 1/2x multiplied by cos 1/2x is...
    cos^2(\frac{x}{2})

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    (Original post by SeanFM)
    Woops! 16Characters was kindly helping you already so I shall leave this one to him.
    Well strictly speaking you were the first to offer any actual help, I was just verifying the question :-)
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    (Original post by 16Characters....)
    cos squared of x/2 (sorry I don't know the latex for cos squared)
    If you wanna know the LaTeX look at my post
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    So at the moment I have for the numerator:

    cos (x/2) + (cos^2 (x/2) + sin^2 (x/2))/2

    and I know that sin^2x + cos^2x = 1,

    but I have sin^2 (x/2) and cos^2 (x/2) instead so what do I do?
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    Damn that was simpler LaTeX than I had expected...
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    (Original post by jordanwu)
    So at the moment I have for the numerator:

    cos (x/2) + (cos^2 (x/2) + sin^2 (x/2))/2

    and I know that sin^2x + cos^2x = 1,

    but I have sin^2 (x/2) and cos^2 (x/2) instead so what do I do?
    You can apply the normal identity. All that  sin^2 \alpha + cos^2 \alpha = 1 means is that the sum of the squares of the sine and cosine of a given angle  \alpha is 1. x/2 is still an angle so the identity holds for that also.
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    (Original post by 16Characters....)
    Damn that was simpler LaTeX than I had expected...
    Hahaha yeah - I usually type what seems sensible and see what happens :lol:

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    (Original post by jordanwu)
    So at the moment I have for the numerator:

    cos (x/2) + (cos^2 (x/2) + sin^2 (x/2))/2

    and I know that sin^2x + cos^2x = 1,

    but I have sin^2 (x/2) and cos^2 (x/2) instead so what do I do?
    Carry on as normal, the identity works as long as the angles are the same.

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    Ok thanks guys I got to the answer.
    So the next part is: find the value of x at which the gradient of the curve is 1/12, giving your answer to 3sf.
    So 1/12=(1/2 + cos (x/2)/(2 + cos (x/2))^2 , but I keep getting stuck on rearranging. Any tips?
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