2
$\begingroup$

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:

$$\tan{X}=\tan{\left(2\arctan{2}+\arcsin{\frac{4}{5}}\right)}=\frac{\tan{(2\arctan{2})}+\tan{\left(\arcsin{\frac{4}{5}}\right)}}{1-\tan{(2\arctan{2})}\cdot\tan{\left(\arcsin{\frac{4}{5}}\right)}}.$$

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

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

ii):

$$M=\tan{\left(\arcsin{\frac{4}{5}}\right)}=\frac{\sin{\left(\arcsin{\frac{4}{5}}\right)}}{\cos{\left(\arcsin{\frac{4}{5}}\right)}}=\frac{\sin{\left(\arcsin{\frac{4}{5}}\right)}}{\sqrt{1-\sin^2{\left(\arcsin{\frac{4}{5}}\right)}}}=\frac{\frac{4}{5}}{\sqrt{1-\left(\frac{4}{5}\right)^2}}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}.$$

So,

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

The book says $X=\pi.$

$\endgroup$
5
  • $\begingroup$ $\pi$ 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
6
$\begingroup$

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 $\pi$.

$\endgroup$
1
  • $\begingroup$ I understand 100%. Thank you buddy! $\endgroup$
    – Parseval
    Sep 16 '17 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.