Author Topic: T formulae question (Read 1346 times) Tweet Share

hinakamishiro · « **on:** July 31, 2016, 10:55:03 am »

Hi guys! I was wondering if I could have some help with this t formulae proof? Thanks!

RuiAce · « **Reply #1 on:** July 31, 2016, 11:19:25 am »

Quote from: hinakamishiro on July 31, 2016, 10:55:03 am

Hi guys! I was wondering if I could have some help with this t formulae proof? Thanks!

$\text{Taking the hint, we wish to prove that }t_1^2=4t_2^2$
$\text{Let }t_1=\tan \frac{x}{2}\text{ and }t_2=\tan \frac{y}{2}\\ \text{Then, using what we are given:}$
$\begin{align*}\frac{1-t_1^2}{1+t_1^2}&=\frac{5\left(\frac{1-t_2^2}{1+t_2^2}\right)-3}{5-3\left(\frac{1-t_2^2}{1+t_2^2}\right)} \quad \left(\times \frac{1+t_2^2}{1+t_2^2}\right)\\ &= \frac{5-5t_2^2-3-3t_2^2}{5+5t_2^2-3+3t_2^2}\\ &= \frac{2-8t_2^2}{8t_2^2+2}\\ &= \frac{1-4t_2^2}{4t_2^2+1}\\ \therefore (1-t_1^2)(4t_2^2+1)&=(1+t_1^2)(1-4t_2^2)\\ 4t_2^2+1-4t_1^2t_2^2-t_1^2&=1-4t_2^2+t_1^2-4t_1^2t_2^2\\ 2t_1^2&=8t_2^2\\ \tan^2 \frac{x}{2}&=4\tan^2 \frac{y}{2} \end{align*}$

hinakamishiro · « **Reply #2 on:** July 31, 2016, 01:03:37 pm »

Quote from: RuiAce on July 31, 2016, 11:19:25 am

$\text{Taking the hint, we wish to prove that }t_1^2=4t_2^2$
$\text{Let }t_1=\tan \frac{x}{2}\text{ and }t_2=\tan \frac{y}{2}\\ \text{Then, using what we are given:}$
$\begin{align*}\frac{1-t_1^2}{1+t_1^2}&=\frac{5\left(\frac{1-t_2^2}{1+t_2^2}\right)-3}{5-3\left(\frac{1-t_2^2}{1+t_2^2}\right)} \quad \left(\times \frac{1+t_2^2}{1+t_2^2}\right)\\ &= \frac{5-5t_2^2-3-3t_2^2}{5+5t_2^2-3+3t_2^2}\\ &= \frac{2-8t_2^2}{8t_2^2+2}\\ &= \frac{1-4t_2^2}{4t_2^2+1}\\ \therefore (1-t_1^2)(4t_2^2+1)&=(1+t_1^2)(1-4t_2^2)\\ 4t_2^2+1-4t_1^2t_2^2-t_1^2&=1-4t_2^2+t_1^2-4t_1^2t_2^2\\ 2t_1^2&=8t_2^2\\ \tan^2 \frac{x}{2}&=4\tan^2 \frac{y}{2} \end{align*}$

Thanks. I have another question for t formulae which is a bit different which I hope I could have some help with as well. Thanks

RuiAce · « **Reply #3 on:** July 31, 2016, 01:12:25 pm »

$\text{Test }x=\pi \text{ separately:} \\ \cos \pi - \sin \pi = -1 \neq 1\\ \text{All good.}$
$\text{Let }t=\tan \frac{x}{2}$
$\begin{align*}\cos x - \sin x &=1\\ \frac{1-t^2}{1+t^2} - \frac{2t}{1+t^2} &= 1 \\ 1-t^2-2t &= 1+t^2\\ -2t-2t^2&=0 \\ t^2+t&=0\end{align*}$
Is there anything in particular that's troubling you? Most of these questions are just substituting what you know and then tackling with algebra.
$\text{Solving: }t(t+1)=0\text{ so }t=0\text{ or }t=-1\\ \therefore \tan \frac{x}{2}=0, \quad \tan \frac{x}{2}=-1$
$\tan \frac{x}{2}=0 \text{ solves to give }\frac{x}{2}=0, \pi \iff x=0, 2\pi\\ \tan \frac{x}{2}=-1\text{ solves to give }\frac{x}{2}=\frac{3\pi}{4} \iff x=\frac{3\pi}{2}$
$\therefore x=0, \frac{3\pi}{2}, 2\pi$

hinakamishiro · « **Reply #4 on:** July 31, 2016, 03:11:06 pm »

Quote

Is there anything in particular that's troubling you? Most of these questions are just substituting what you know and then tackling with algebra

Thats all thanks, i think it just kinda didn't click the first time. Thanks for your help!

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