Quick double angle query

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  1. SubAtomic's Avatar
    1 24-07-2012  17:58
    • Exalted and Worshipped Member
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    Quick double angle query
    Hi, so I am studying something on quadratic curves, the book doesn't ask me to do it but I like to know where they go from one formula to another. If I was a cat I would be dead

    So just a quick confirmation really,

    \displaystyle \tan (2 \theta)= \dfrac{B}{A-C}


    Can be obtained by rearranging and manipulating

    \displaystyle (C-A) \sin (2 \theta)+ B \cos(2 \theta)=0

    Tried it but I got as far as

    \displaystyle \dfrac{ \sin (2 \theta)}{ \cos (2 \theta)}= \dfrac{-B}{C-A}

    So I assume that

    \displaystyle \dfrac{-B}{C-A}=\dfrac{B}{A-C}


    or would that be a wrong assumption?

    Ignore me I just subbed some numbers in and can see it seems to work
    Last edited by SubAtomic; 24-07-2012 at 18:01.
  2. nuodai's Avatar
    2 24-07-2012  18:02
    • PS Helper
    • TSR Legend
    Re: Quick double angle query
    Your assumption is correct. It's because C-A=-(A-C), and so the minus signs cancel.
  3. TenOfThem's Avatar
    3 24-07-2012  18:04
    • TSR Royalty
    Re: Quick double angle query
    (Original post by SubAtomic)

    So I assume that

    \displaystyle \dfrac{-B}{C-A}=\dfrac{B}{A-C}

    You know that you can multiply the numerator and denominator of a fraction by the same number and the fraction remains the same ... yes?


    In this case you have multiplied by -1
  4. SubAtomic's Avatar
    4 24-07-2012  18:10
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    Re: Quick double angle query
    (Original post by TenOfThem)
    You know that you can multiply the numerator and denominator of a fraction by the same number and the fraction remains the same ... yes?


    In this case you have multiplied by -1
    Yep, one thing goes in one thing escapes, hopefully am swapping for better things though. Don't know what I was thinking I like reassurance too much. Bit fried from all the A' B' C' u^2 uv v^2 sincos sin^2 cos^2 etc etc

    Thank you both for clarifying:cool:
    Last edited by SubAtomic; 24-07-2012 at 18:15.
  5. SubAtomic's Avatar
    5 20-08-2012  15:01
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    Re: Quick double angle query
    Not double angle I know but how does this work, don't get how the last line is equal to the second.

    \displaystyle \dfrac{1-\cos \theta}{ \theta} \\ \\ = \dfrac{2 \sin^2( \frac{1}{2} \theta)}{ \theta} \\ \\ = \left( \dfrac{\sin (\frac{1}{2} \theta)}{ (\frac{1}{2} \theta)}\right)^2 \times \left(\frac{1}{2} \theta \right)

    So how do I go about multiplying out this as this is where the problem lies I think

    \displaystyle \left( \dfrac{\sin (\frac{1}{2} \theta)}{ (\frac{1}{2} \theta)}\right) \times \left( \dfrac{\sin (\frac{1}{2} \theta)}{ (\frac{1}{2} \theta)}\right)\times \left(\frac{1}{2} \theta \right)


    All I can think is I'd end up with

    \displaystyle \dfrac {(\sin ( \frac{1}{2} \theta))^2}{ (\frac{1}{2} \theta)^2} \times \left( \frac {1}{2} \theta \right)

    What do I do with the half theta squared? Does it become a quarter theta? And what do I do when multiplying by the half theta?


    Back later, any help appreciated:cool:
    Last edited by SubAtomic; 20-08-2012 at 15:10.
  6. Lord of the Flies's Avatar
    6 20-08-2012  15:20
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: Quick double angle query
    (Original post by SubAtomic)
    ...
    \displaystyle \left( \dfrac{\sin (\frac{1}{2} \theta)}{ (\frac{1}{2} \theta)}\right)^2 \times \left(\frac{\theta}{2} \right)= \dfrac{\sin^2 (\frac{1}{2} \theta)}{ \frac{1}{4} \theta^2} \times \left(\frac{\theta}{2} \right)=\dfrac{\theta\cdot \sin^2 (\frac{1}{2} \theta)}{ \frac{1}{2} \theta^2} =\dfrac{2 \sin^2( \frac{1}{2} \theta)}{ \theta}
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  7. SubAtomic's Avatar
    7 20-08-2012  15:23
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    Re: Quick double angle query
    (Original post by Lord of the Flies)
    \displaystyle \left( \dfrac{\sin (\frac{1}{2} \theta)}{ (\frac{1}{2} \theta)}\right)^2 \times \left(\frac{\theta}{2} \right)= \dfrac{\sin^2 (\frac{1}{2} \theta)}{ \frac{1}{4} \theta^2} \times \left(\frac{\theta}{2} \right)=\dfrac{\theta\cdot \sin^2 (\frac{1}{2} \theta)}{ \frac{1}{2} \theta^2} =\dfrac{2 \sin^2( \frac{1}{2} \theta)}{ \theta}
    Just thought about what I did and didn't square the theta in the denom for some reason, I need to turn the theta into x for a while until it second nature, as in can just leave it as theta.

    Thanks:cool:
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