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Math Help - help me prove it

  1. #1
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    Smile help me prove it

    hello
    let \left ( a,x \right )\in \mathbb{R}^{2} such that ax\neq 1,prove the following ;
    arctan(x)+arctan(a)=arctan(\frac{a+x}{1-ax})+\epsilon \pi
    ( \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:

    \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 \epsilon\in\mathbb Z


    Hint #2:

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

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


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

    \arctan(x)+\arctan(y)\:=\:\arctan\left(\frac{x+y}{  1-xy}\right)+n\pi, \quad<br />
n \in\{-1,0,1\}
    Let: . \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: . \arctan(x) + \arctan(y) \;=\;\alpha + \beta

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

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

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

    Therefore: . \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 +n\pi,n\in Z because tan(x-\pi)=tan(x+\pi)=tan(x).
    right thanks a lot.
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