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    (Original post by aymanzayedmannan)
    Not a STEP-er but I'm joining this thread to pick up on useful techniques. Here's an easy one that might be helpful:

    Factorising x^{4} + 1 and polynomials of the like -

    It's fairly easy to write out \displaystyle x^{4} + 1 \equiv \left ( x^{2} + \alpha x +1 \right )\left( x^{2}+\beta x +1 \right ) and then compare coefficients using C3 techniques, but in case you're feeling adventurous,



    This could come in handy in certain integrals. Try the classic \displaystyle \int \sqrt{\tan x} \ \mathrm{d}x using this.
    I gave a fair amount of effort to do this, so please hope it's correct!!!
    P.s sorry for my messy writing, I've been ill for the past few days so don't have the steadiest hand ever
    (also final answer is at the top)
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    There is a very clever way of doing this, but I do not quite remember it.
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    (Original post by EnglishMuon)
    I gave a fair amount of effort to do this, so please hope it's correct!!!
    The answer looks correct at a first glance. If you're interested in a slightly neater way of doing this, check here.
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    (Original post by DylanJ42)
    this post is more useful i think, and im surprised it hasnt been posted about yet; (maybe everyone knows it already)

     \displaystyle \left(a^n + b^n \right) = \left(a-b\right)\left(a^{n-1} - a^{n-2}b +  a^{n-3}b^2 - a^{n-4}b^3 + ... - ab^{n-2} + b^{n-1}\right)

    ^^ this is only valid when n is even. Notice how you alternate signs the whole way down the right hand side bracket, starting and ending with plus signs

     \displaystyle \left(a^n - b^n\right) = \left(a-b\right)\left(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + ab^{n-2} + b^{n-1}\right)

    valid for all integer n, and notice how the right hand bracket is all plus signs
    very useful factorisation, but one of your signs is wrong on the first one, and it works for odd n, not even
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    (Original post by StrangeBanana)
    very useful factorisation, but one of your signs is wrong on the first one, and it works for odd n, not even
    i see another mistake too, thank you for pointing that out :facepalm:
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    (Original post by Zacken)
    The answer looks correct at a first glance. If you're interested in a slightly neater way of doing this, check here.
    lol thanks! I was on the verge of anal leakage until you said you didn't come up with the technique yourself but nevertheless a great solution!
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    (Original post by EnglishMuon)
    I gave a fair amount of effort to do this, so please hope it's correct!!!
    P.s sorry for my messy writing, I've been ill for the past few days so don't have the steadiest hand ever
    (also final answer is at the top)
    The answer is spot on with this method. But please do check out the elegant solution Zacken posted!
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    (Original post by aymanzayedmannan)
    The answer is spot on with this method. But please do check out the elegant solution Zacken posted!
    Thanks! (phew thats a relief!) Yeah Ive thought about this (I+J)+(I-J) method before but forgot to consider it here. Makes sense though but Im not sure how long it wouldve taken me to choose  \sqrt{cotx} for J, just something I would have to look ahead for I suppose. Thanks for the help
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    Is it worth mentioning T substitutions?
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    (Original post by mpaterson)
    Is it worth mentioning T substitutions?
    As in the tangent sub? I think it's already been mentioned.
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    (Original post by aymanzayedmannan)
    As in the tangent sub? I think it's already been mentioned.
    hmm I was thinking:
    

\displaystyle t = \tan \frac{ \theta }{2}



\displaystyle \cos \theta = \frac{ 1 - t^2 }{ 1 + t^2}



\displaystyle \sin \theta = \frac{ 2t }{ 1 + t^2}



\displaystyle \frac{d\theta}{dt} = \frac{2}{1 + t^2}
    couldn't see that anywhere earlier

    p.s. anyone know how to space lines in latex?
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    (Original post by mpaterson)
    hmm I was thinking:
    

\displaystyle t = \tan \frac{ \theta }{2}



\displaystyle \cos \theta = \frac{ 1 - t^2 }{ 1 + t^2}



\displaystyle \sin \theta = \frac{ 2t }{ 1 + t^2}



\displaystyle \frac{d\theta}{dt} = \frac{2}{1 + t^2}
    couldn't see that anywhere earlier

    p.s. anyone know how to space lines in latex?
    A link had been posted by shamika, but you can definitely expand on it.

    I just do

    [tex] ... [/ tex]

    [tex] next line [/ tex]

    EDIT: it's under number 8 along with a wiki link.
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    (Original post by aymanzayedmannan)
    A link had been posted by shamika, but you can definitely expand on it.

    I just do

    [tex] ... [/ tex]

    [tex] next line [/ tex]

    EDIT: it's under number 8 along with a wiki link.
    Ahh I see, it's obvious now.

    Thanks!
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    (Original post by mpaterson)
    hmm I was thinking:
    

\displaystyle t = \tan \frac{ \theta }{2}



\displaystyle \cos \theta = \frac{ 1 - t^2 }{ 1 + t^2}



\displaystyle \sin \theta = \frac{ 2t }{ 1 + t^2}



\displaystyle \frac{d\theta}{dt} = \frac{2}{1 + t^2}
    couldn't see that anywhere earlier

    p.s. anyone know how to space lines in latex?
    You should Use these largely when there trig in the denominator I think and denominator and numerator. But alot of the times you will instructed or hinted at through a first part!


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    Roots of unity

    If
    

\omega
    is the nth root of unity,

    we can write

    

(\omega)^{n} - 1 = 0

(\omega - 1)(\omega^{n-1} + \omega^{n-2} ... + 1) = 0

    Since (normally) the first bracket isn't zero, you have the somewhat useful result that the latter is zero. The geometrical implication of this (by considering an n-sided polygon in the Argand diagram, centred at the origin) is that the sum of the vectors from the centre of a polygon to its individual vertices is zero.

    See III/97/3, I/90/2.
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    (Original post by mpaterson)
    p.s. anyone know how to space lines in latex?
    You can also use backslashes, as follows to leave a blank line.
    So:
    [latex]a \\
    b[/latex]
    becomes:
    a \\

b
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    (Original post by Krollo)
    Roots of unity

    If
    

\omega
    is the nth root of unity,

    we can write

    

(\omega)^{n} - 1 = 0

(\omega - 1)(\omega^{n-1} + \omega^{n-2} ... + 1) = 0

    Since (normally) the first bracket isn't zero, you have the somewhat useful result that the latter is zero. The geometrical implication of this (by considering an n-sided polygon in the Argand diagram, centred at the origin) is that the sum of the vectors from the centre of a polygon to its individual vertices is zero.

    See III/97/3, I/90/2.
    In addition to this, it's useful to know that if z is a root, then so is \bar{z} and that, for any root z

    \displaystyle \frac{1}{z}=\bar{z}
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    (Original post by Student403)
    :toofunny:
    But anyway how you been my friend?
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    (Original post by EnglishMuon)
    on the verge of anal leakage
    wtf man
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    (Original post by M14B)
    There is a very clever way of doing this, but I do not quite remember it.
    There is but you kind of have to know what you're looking for or seen it before or something. I think it would be difficult to see if you'd never seen it done before or seen something similar.
 
 
 
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