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Find value of $2A$ if $A=\frac{3\tan\left(A\right)}{1-\tan\left(A\right)}-1$

My first thought was rewriting the $1$ as $\frac{1-\tan(A)}{1-\tan(A)}$, which implies that $A=\frac{4\tan\left(A\right)-1}{1-\tan\left(A\right)}$

I then noticed that it awfully mirrored the tangent angle addition identity, $\tan\left(\alpha+\beta\right)=\frac{\tan\left(\alpha\right)+\tan\left(\beta\right)}{1-\tan\left(\alpha\right)\tan\left(\beta\right)}$.

However, I am quite unsure on how to progress from here, or if my current progress is even in the correct direction. Any assistance would be greatly appreciated!

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  • $\begingroup$ I suspect there be solutions like $q\pi$ where $q\in\Bbb Q$. $\endgroup$ Commented Apr 8 at 18:50
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    $\begingroup$ The solutions assymptotically will lie on the assymptotes of $y=\frac{3\tan(x)}{1-tan(x)}-1$, i.e. when $x=\frac{1}{4}(4\pi n+n)$. Not sure if a closed form solution can be found. $\endgroup$
    – Shean
    Commented Apr 8 at 19:54
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    $\begingroup$ @Shean : the asymptotes are in $\pi/4+n \pi$ $\endgroup$
    – Jean Marie
    Commented Apr 8 at 22:07
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    $\begingroup$ This problem has an infinite number of solutions... $\endgroup$
    – Jean Marie
    Commented Apr 8 at 22:09
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    $\begingroup$ It's almost like the tangent double angle formula $\tan 2A = \dfrac {2 \tan A}{1-\tan^2 A}$... $\endgroup$
    – bjcolby15
    Commented Apr 9 at 22:18

2 Answers 2

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To get rid of the discontinuities, multiply everything by $(1-\tan(A))\cos(A)$ which makes that we look for the zero's of function $$f(A)=(A+4) \sin (A)-(A+1) \cos (A)$$

As said in comments, the solutions will be closer and closer to $\left(n+\frac{1}{4}\right)\pi$. So, let $$A= \left(n+\frac{1}{4}\right)\pi+x$$ and expand to obtain $$6\cos(x)+(4x+(4n+1)\pi+10)\sin(x)=0$$

Now, using the very first terms of the Taylor expansions of the sine and cosine functions around $x=0$, you will have $$6+((4n+1)\pi+10) x+O(x^2)=0 \quad \implies x=-\frac 6{(4n+1)\pi+10 }$$ that is to say $$A_{(n)} \sim \left(n+\frac{1}{4}\right)\pi-\frac 6{(4n+1)\pi+10}$$

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Let us start by inverting the homographic relationship :

$$A=\frac{4 \tan A -1}{1- \tan A} \ \iff \ \tan A=\frac{A+1}{A+4}=1-\frac{3}{A+4}$$

The idea is to use the following expansion of the "principal branch" of $\arctan(1-z)$ :

$$\arctan(1-z)=\frac{\pi}{4}\color{red}{-\frac{z}{2}-\frac{z^2}{4}-\frac{z^3}{12}}+\frac{z^5}{40}+\frac{z^6}{48}+\frac{z^7}{112}-\frac{z^9}{288}+\cdots$$

(the first terms in red will be those used in the sequel ; please note a certain irregularity of the expansion : sign changes, missing terms in $z^{4k}$).

... and taking into account the fact that the solution is close to $(n+\tfrac14)\pi$, we can write the solutions under the approximate form : $$A\approx \color{blue}{n\pi} + \frac{\pi}{4} -\frac{z}{2}-\frac{z^2}{4}-\frac{z^3}{12} \ \ \text{with} \ \ z=\frac{3}{(n+\tfrac14)\pi+4}$$

(the term $n\pi$ in blue has been added in order to take into account the multivalued property of "arctan" function).

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