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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 :wink:

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 :colondollar:

(edited 11 years ago)

Reply 1

nuodai

Your assumption is correct. It's because

C-A=-(A-C)

, and so the minus signs cancel.

Reply 2

TenOfThem

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

Reply 3

SubAtomic

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 :colondollar:

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:

(edited 11 years ago)

Reply 4

SubAtomic

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:

(edited 11 years ago)

Reply 5

Lord of the Flies

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}

Reply 6

SubAtomic

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: