Proceeding as you guys have tought me on my previous posts by setting the whole thing equal to $X$ taking the tangent of the whole expression and then using the additon formula for tangent:


So lets split this into two sub-computations like this:

i): $$N=\tan{(2\arctan{2})}=\tan{(\arctan{2}+\arctan{2})}=\frac{2\tan{(\arctan{2})}}{1-\tan^2{(\arctan{2})}}=\frac{2\cdot 2}{1-2^2}=-\frac{4}{3}.$$




$$\tan{X}=\frac{N+M}{1-MN}=\frac{-\frac{4}{3}+\frac{4}{3}}{1+\frac{4}{3}\cdot\frac{4}{3}}=0\Longrightarrow X=\arctan{0}=0.$$

The book says $X=\pi.$

  • $\begingroup$ $\pi$ is the right answer $\endgroup$ Sep 16 '17 at 21:56
  • 1
    $\begingroup$ Solutions of the equation $\tan x = c$ are determined only up to addition by integer multiples of $\pi$. Moreover, without any further input, all of them equally makes sense. In your case, you may write $$ X = \pi - 2\arctan\left(\frac{1}{2}\right) + \arctan\left(\frac{4}{3}\right) $$ and then proceed. $\endgroup$ Sep 16 '17 at 21:57
  • $\begingroup$ @SangchulLee - I'm still not sure where my solution is wrong. $\endgroup$
    – Parseval
    Sep 16 '17 at 21:59
  • 1
    $\begingroup$ Your mistake is the part where you concluded $X = 0$ from $\tan X = 0$. As I explained, $\tan X = 0$ only tells you that $X = n \pi$ for some $n \in \mathbb{Z}$. You still need to determine which $n$ fits into your $X$. This can be done if you carefully examine $X$. For instance, Chappers explains one possible way. $\endgroup$ Sep 16 '17 at 22:00
  • $\begingroup$ Yes, I forgot to add the periodicity of tangent function. $\endgroup$
    – Parseval
    Sep 16 '17 at 22:08

The last implication is false: $\tan{X} = 0 \implies X = k\pi$ for some $k \in \mathbb{Z}$. To find out which value, it's probably simplest to estimate the terms:

  • $\arctan{2}>\arctan{1}=\pi/4$, so the sum is certainly bigger than $\pi/2$.
  • On the other hand, $0<\arcsin{(4/5)} < \pi/2$ and $\arctan{2}<\pi/2$, so the result is bounded above by $3\pi/2$.

The only multiple of $\pi$ between these is $\pi$ itself, so the value of the sum must be $\pi$.

  • $\begingroup$ I understand 100%. Thank you buddy! $\endgroup$
    – Parseval
    Sep 16 '17 at 22:08

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