Compute $2\arctan{2}+\arcsin{\frac{4}{5}}.$

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 $X=\pi.$

• $\pi$ is the right answer Sep 16 '17 at 21:56
• 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. Sep 16 '17 at 21:57
• @SangchulLee - I'm still not sure where my solution is wrong. Sep 16 '17 at 21:59
• 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. Sep 16 '17 at 22:00
• Yes, I forgot to add the periodicity of tangent function. 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 $\pi$.