# Find $x$ such that $\arctan(3/2)+\dots=\arctan x$

Find $x$ such that

$$\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3) = \arctan x.$$

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You should leave the question here even if it has been solved. That way it may help other people with the same or similar questions in the future! –  Deven Ware Nov 21 '12 at 23:01
To add on Deven Ware's comment, if one of the answers helped you to reach a solution you should accept it by click on the check mark under the vote up/down arrows; and if none of the answers helped, you should post your own answer and accept it (although you have to wait for two days before accepting your own answer). –  Asaf Karagila Nov 21 '12 at 23:22

You may use the fact if $x > 0$, the argument of the complex number $x+iy$ is $\tan^{-1} \left(\frac{x}{y}\right)$. So if you compute the product

$$(2+3i)(4+5i)(2-5i)(3-8i) = 920 - 531i$$

you get that

$$\tan^{-1}\left(\frac{3}{2}\right) + \tan^{-1}\left(\frac{5}{4}\right) + \tan^{-1}\left(-\frac{5}{2}\right) + \tan^{-1}\left(-\frac{8}{3}\right) \equiv \tan^{-1}\left(-\frac{531}{920}\right) \mod 2\pi$$

By computing both side on a calculator, you can check this actually an equality.

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Yes, it is possible that the left hand side of the expression is not in the range of arctan. Good catch... –  Thomas Andrews Nov 21 '12 at 22:07

Hint: A repetition of applying the formula

$$\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}$$

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While that does result in the answer, why is that procedure motivated? As in, what does the student gain from this exercise other than the repetition of a formula? –  000 Nov 21 '12 at 21:08
@Limitless Amongst other things, the student learns how to do certain types of logical operations, and learns that this sort of expression is necessarily rational. There is a lot to be learned from working out these sorts of problems by hand. –  Thomas Andrews Nov 21 '12 at 21:26
@Limitless: If one does not work out mathematics problems in details, he will not learn mathematics. –  Mhenni Benghorbal Nov 22 '12 at 2:02

Take $\tan$ of both sides, and repeatedly apply the compound angle formula.

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I don't see how there's motivation in the closed form. This strikes me as the type of problem best left to Wolfram Alpha, just as one leaves the integration of $\frac{x^5+x}{x^3+x^2+1}$ to Wolfram Alpha. Do you see how or why this equation would be a motivated exercise? –  000 Nov 21 '12 at 21:05
I can't think of a reason, other than learning the formula for $\tan(A+B)$ –  Daniel Littlewood Nov 21 '12 at 21:21
@DanielLittlewood I think it is generally useful to learn not just such formulas, but how to apply them. –  Thomas Andrews Nov 21 '12 at 21:39
I agree, but I didn't set the exercise :-) –  Daniel Littlewood Nov 21 '12 at 21:43

Let $\omega_1= 2+3i, \omega_2=4+5i, \omega_3=2-5i , \omega_4=3-8i$. Then

$$\arg(\omega_1)= \arctan(\frac{3}{2}), ...$$

Finc $\omega_1 \omega_2 \omega_3 \omega_4 =a+bi$ and then

$$\frac{b}{a}= \arg(\omega_1 \omega_2 \omega_3 \omega_4 ) =\arctan(x) \,.$$

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\begin{align} & {} \quad \tan(a+b+c+d) \\ & = \frac{\tan a+\tan b+\tan c+\tan d\ \overbrace{ - \tan a\tan b\tan c - \cdots}^\text{4 terms}}{1-\ \underbrace{\tan a \tan b- \cdots}_\text{6 terms} +\tan a\tan b\tan c\tan d} \end{align}

Therefore \begin{align} & {} \quad \tan\Big(\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3)\Big) \\ & = \frac{\frac32 + \frac54 + \frac{-5}2 +\frac{-8}3 - (\text{sum of four terms, each a product of three numbers})}{1 - (\text{sum of six terms, each a product of two numbers}) + (\text{a product of four numbers}))} \end{align}

You get a number. Then you say $\tan x=\text{that number}$, and find the tangent.

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