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    A while ago some of you helped me do the first part of this question:

    Given that t=tan(x/2), prove that cos x = (1-t^2)/(1+t^2)

    However, I am still stuck on a subsequent part:

    Therefore prove that dx/dt = 2/(1+t^2)

    Any ideas? Sure someone, especially Fermat will know! Thanks
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    (Original post by Hoofbeat)
    A while ago some of you helped me do the first part of this question:

    Given that t=tan(x/2), prove that cos x = (1-t^2)/(1+t^2)

    However, I am still stuck on a subsequent part:

    Therefore prove that dx/dt = 2/(1+t^2)

    Any ideas? Sure someone, especially Fermat will know! Thanks

    I would approach this question as follows:


    start with cos x = (1-t^2)/(1+t^2) and differentiate implicitly

    hence, -sinx.dx/dt = (1+t^2)(-2t)-(1-t^2)(2t)/(1+t^2)^2

    then find an expression for sinx in terms of t, which turns out to be sinx=sqrt2t/(1+t^2)

    sub into the exrpression, so dx/dt=-((1+t^2)(-2t)-(1-t^2)(2t))sqrt2t/(1+t^2)^3

    and simplify.
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    I am very surprised that tan(x/2) substitutions are in P3 but anyway...

    t = tan(x/2)
    dt/dx = (1/2)sec^2(x/2)
    dx/dt = 2cos^2(x/2)

    Now, the way I like to do this is to consider that if tan(x/2) = t then we have a right angled triangle with an angle x/2 where the opposite side is t and the adjacent side is 1. By pythagoras' theorem the hypotenuse side is sqrt(1+t^2). Therefore,

    cos(x/2) = ADJ/HYP = 1/sqrt(1+t^2)

    Plug this in:

    dx/dt = 2cos^2(x/2) = 2( 1/sqrt(1+t^2) )^2 = 2/(1+t^2)
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    (Original post by mikesgt2)
    I am very surprised that tan(x/2) substitutions are in P3 but anyway...

    t = tan(x/2)
    dt/dx = (1/2)sec^2(x/2)
    dx/dt = 2cos^2(x/2)

    Now, the way I like to do this is to consider that if tan(x/2) = t then we have a right angled triangle with an angle x/2 where the opposite side is t and the adjacent side is 1. By pythagoras' theorem the hypotenuse side is sqrt(1+t^2). Therefore,

    cos(x/2) = OPP/ADJ = 1/sqrt(1+t^2)

    Plug this in:

    dx/dt = 2cos^2(x/2) = 2( 1/sqrt(1+t^2) )^2 = 2/(1+t^2)
    Thanks I'm suprised by it too, as I thought it was P4! It's from a review exericse so the questions are usually harder than questions from papers! Thats what i'm hoping anyways...
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    (Original post by mikesgt2)
    cos(x/2) = OPP/ADJ = 1/sqrt(1+t^2)
    But cos = adj/hyp!!!

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    I've solved it:

    Opposite = t
    Adjacent = 1

    therefore, cos(x/2) = 1/sqrt(t^2+1)

    therefore, 2cos^2(x/2) = 2/(t^2+1)

    so dx/dt=2/(t^2+1) as required!

    Thanks everyone for your help!!!
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    (Original post by Hoofbeat)
    But cos = adj/hyp!!!

    Sorry, my mistake . I will edit the post.
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    (Original post by Hoofbeat)
    Thanks I'm suprised by it too, as I thought it was P4! It's from a review exericse so the questions are usually harder than questions from papers! Thats what i'm hoping anyways...
    Yeah, tan half angle substitutions will not be on P3 exams, not even on P4 as far as I am aware. I have seen such questions in the P3 and 4 review exercises but it is introdcued properly in P5.
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    (Original post by Hoofbeat)
    A while ago some of you helped me do the first part of this question:

    Given that t=tan(x/2), prove that cos x = (1-t^2)/(1+t^2)

    However, I am still stuck on a subsequent part:

    Therefore prove that dx/dt = 2/(1+t^2)

    Any ideas? Sure someone, especially Fermat will know! Thanks
    Well, sometimes, but not always
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    (Original post by mikesgt2)
    ...

    Now, the way I like to do this is to consider that if tan(x/2) = t then we have a right angled triangle with an angle x/2 where the opposite side is t and the adjacent side is 1. By pythagoras' theorem the hypotenuse side is sqrt(1+t^2).

    ...
    I use a rt-angled triangle as well when I'm dealing with tan of half-angles, t=tan(x/2).
    But I take the opposite as 2t, the adjacent as 1-t², and pythagoras gives the hypotenuse as 1+ t²

    Then,

    sinø = 2t/(1+t²
    cosø = (1-t²)/(1+t²)
    tanø = 2t/(1-t²)

    where t=tan(ø/2)

    which are the definitions for the half-angle identities anyway. I find it easier to remember them (i.e work them out) doing it like that.
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    (Original post by Hoofbeat)
    Thanks I'm suprised by it too, as I thought it was P4! It's from a review exericse so the questions are usually harder than questions from papers! Thats what i'm hoping anyways...
    Only marginally.
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    (Original post by bono)
    Only marginally.
    Don't burst my bubble!!!

    No seriously, I agree that most of them are of the same standard, but the occaisional ones are an awful lot harder! lol.

    Anyways, must go, am going to do an M2 paper!
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    Hey to anyone out there struggling with differentiation or finding it hard tocheck answers correctly then i found a site which has various programs including one on differentiation that will differentiate for you, using many different ways. Its not on the site as i emailed the siteowner directly and he sent me the program. however if u want it jus do the same.....

    The site is www.geocities.com/softpro101
 
 
 

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