SubAtomic
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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
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nuodai
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Your assumption is correct. It's because C-A=-(A-C), and so the minus signs cancel.
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TenOfThem
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(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
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SubAtomic
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(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:
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SubAtomic
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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:
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Lord of the Flies
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(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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SubAtomic
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(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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