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Thread: help me prove it

  1. #1
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    Smile help me prove it

    hello
    let $\displaystyle \left ( a,x \right )\in \mathbb{R}^{2}$ such that $\displaystyle ax\neq 1$,prove the following ;
    $\displaystyle arctan(x)+arctan(a)=arctan(\frac{a+x}{1-ax})+\epsilon \pi $
    ( $\displaystyle \epsilon \in \left \{ 1,-1,0 \right \}$ )
    just hints guys,thanks a lot.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Hint #1:

    $\displaystyle \frac{\tan(\arctan a)+\tan(\arctan x)}{1-\tan(\arctan a)\tan(\arctan x)}\ =\ \frac{a+x}{1-ax}\ =\ \frac{\tan\left[\arctan\left(\dfrac{a+x}{1-ax}\right)\right]+\tan(\epsilon\pi)}{1-\tan\left[\arctan\left(\dfrac{a+x}{1-ax}\right)\right]\tan(\epsilon\pi)}$ where $\displaystyle \epsilon\in\mathbb Z$


    Hint #2:

    $\displaystyle -\frac\pi2\,<\,\arctan\left(\frac{a+x}{1-ax}\right)\,<\,\frac\pi2$ and $\displaystyle -\pi\,<\,\arctan a+\arctan x\,<\,\pi$ (this is to get the range of values of $\displaystyle \epsilon)$
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  3. #3
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    Hello, Raoh!

    A slight variation of The Abstractionist's proof . . .


    Let $\displaystyle \left(x,y \right)\in \mathbb{R}^{2}$ such that $\displaystyle xy \neq 1$, prove the following:

    $\displaystyle \arctan(x)+\arctan(y)\:=\:\arctan\left(\frac{x+y}{ 1-xy}\right)+n\pi, \quad
    n \in\{-1,0,1\}$
    Let: .$\displaystyle \begin{array}{ccccc}\alpha \:=\:\arctan(x) &\Rightarrow& x \:=\:\tan\alpha & {\color{blue}[1]}\\ \beta \:=\:\arctan(y) &\Rightarrow& y \:=\:\tan\beta & {\color{blue}[2]}\end{array}$


    The left side is: .$\displaystyle \arctan(x) + \arctan(y) \;=\;\alpha + \beta$

    Take the tangent: .$\displaystyle \tan(\alpha + \beta) \;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} $

    Substitute $\displaystyle {\color{blue}[1]}$ and $\displaystyle {\color{blue}[2]}$: .$\displaystyle \tan(\alpha + \beta) \;=\;\frac{x+y}{1-xy} $

    Hence: .$\displaystyle \alpha + \beta \;=\;\arctan\left(\frac{x+y}{1-xy}\right) + n\pi$

    Therefore: .$\displaystyle \arctan(x) + \arctan(y) \;=\;\arctan\left(\frac{x+y}{1-xy}\right) + n\pi$

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  4. #4
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    Smile

    another small question guys
    we have $\displaystyle +n\pi,n\in Z$ because $\displaystyle tan(x-\pi)=tan(x+\pi)=tan(x)$.
    right thanks a lot.
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