Question Number 79124 by ~blr237~ last updated on 22/Jan/20
$${Prove}\:{that}\: \\ $$$$\:\mathrm{16}{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{4}{arctan}\left(\frac{\mathrm{1}}{\mathrm{239}}\right)=\pi \\ $$$$ \\ $$
Commented by mind is power last updated on 23/Jan/20
$${nice}\:{one} \\ $$
Answered by ~blr237~ last updated on 27/Jan/20
![let take z=a+ib with a≠0 we can prove that arz≡arctan((b/a))[2π] let named A=4arctan((1/5))−arctan((1/(239))) A≡4arg(5+i)−arg(239+i)[2π] A≡arg[(5+i)^4 ]−arg[(239+i)] [2π] A≡arg[((476+480i)]−arg(239+i)[2π] A≡arg[(((476+480i)(239−i))/(239^2 −1))] [2π] A≡arg[((476×239+480+i(480×239−476))/(238×240))] [2π] A≡arg[((114244+114244i)/(57120)) ] [2π] A≡arg[((114244)/(57120))(1+i)] [2π] A≡0+arg(1+i)[2π] A≡(π/4)[2π] such as arctan((1/(239))),arctan((1/5))∈[0,(π/2)] we have after framing A : −(π/2)≤A≤2π So finaly A=(π/4)](https://www.tinkutara.com/question/Q79711.png)
$${let}\:{take}\:{z}={a}+{ib}\:\:{with}\:{a}\neq\mathrm{0} \\ $$$${we}\:{can}\:{prove}\:{that}\:{arz}\equiv{arctan}\left(\frac{{b}}{{a}}\right)\left[\mathrm{2}\pi\right] \\ $$$${let}\:{named}\:\:{A}=\mathrm{4}{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{239}}\right) \\ $$$${A}\equiv\mathrm{4}{arg}\left(\mathrm{5}+{i}\right)−{arg}\left(\mathrm{239}+{i}\right)\left[\mathrm{2}\pi\right] \\ $$$${A}\equiv{arg}\left[\left(\mathrm{5}+{i}\right)^{\mathrm{4}} \right]−{arg}\left[\left(\mathrm{239}+{i}\right)\right]\:\left[\mathrm{2}\pi\right]\:\:\: \\ $$$${A}\equiv{arg}\left[\left(\left(\mathrm{476}+\mathrm{480}{i}\right)\right]−{arg}\left(\mathrm{239}+{i}\right)\left[\mathrm{2}\pi\right]\right. \\ $$$$\:{A}\equiv{arg}\left[\frac{\left(\mathrm{476}+\mathrm{480}{i}\right)\left(\mathrm{239}−{i}\right)}{\mathrm{239}^{\mathrm{2}} −\mathrm{1}}\right]\:\left[\mathrm{2}\pi\right]\: \\ $$$${A}\equiv{arg}\left[\frac{\mathrm{476}×\mathrm{239}+\mathrm{480}+{i}\left(\mathrm{480}×\mathrm{239}−\mathrm{476}\right)}{\mathrm{238}×\mathrm{240}}\right]\:\left[\mathrm{2}\pi\right] \\ $$$${A}\equiv{arg}\left[\frac{\mathrm{114244}+\mathrm{114244}{i}}{\mathrm{57120}}\:\right]\:\left[\mathrm{2}\pi\right]\: \\ $$$${A}\equiv{arg}\left[\frac{\mathrm{114244}}{\mathrm{57120}}\left(\mathrm{1}+{i}\right)\right]\:\left[\mathrm{2}\pi\right] \\ $$$${A}\equiv\mathrm{0}+{arg}\left(\mathrm{1}+{i}\right)\left[\mathrm{2}\pi\right] \\ $$$${A}\equiv\frac{\pi}{\mathrm{4}}\left[\mathrm{2}\pi\right]\: \\ $$$${such}\:{as}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{239}}\right),{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:\:{we}\:{have}\:{after}\:{framing}\:{A}\::\:−\frac{\pi}{\mathrm{2}}\leqslant{A}\leqslant\mathrm{2}\pi \\ $$$${So}\:\:{finaly}\:\:{A}=\frac{\pi}{\mathrm{4}} \\ $$