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Question Number 17068 by chux last updated on 30/Jun/17
prove that 4tan^(−1) ((1/5))−tan^(−1) ((1/(239))) =π/4
$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$ \\ $$$$\mathrm{4tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right)\:=\pi/\mathrm{4} \\ $$
Answered by ajfour last updated on 30/Jun/17
z_1 =5+i ; z_2 =239+i Arg(z_1 )=tan^(−1) (1/5)=α Arg(z_2 )=tan^(−1) (1/239)=β let z=(z_1 ^4 /z_2 ), then Arg(z)=4α−β z=(z_1 ^4 /z_2 )=(((5+i)^4 )/(239+i))=(((24+10i)^2 (239−i))/(239^2 +1)) =(((476+480i)(239−i))/(239^2 +1)) =(((476×239+480)+i(480×239−476))/(239^2 +1)) z=((114244)/(239^2 +1)) (1+i) z has argument θ=π/4 ⇒ 4α−β=(π/4)+kπ or 4tan^(−1) ((1/5))−tan^(−1) ((1/(239)))=(π/4)+kπ.
$$\mathrm{z}_{\mathrm{1}} =\mathrm{5}+\mathrm{i}\:\:;\:\:\mathrm{z}_{\mathrm{2}} =\mathrm{239}+\mathrm{i} \\ $$$$\mathrm{Arg}\left(\mathrm{z}_{\mathrm{1}} \right)=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}/\mathrm{5}\right)=\alpha \\ $$$$\mathrm{Arg}\left(\mathrm{z}_{\mathrm{2}} \right)=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}/\mathrm{239}\right)=\beta \\ $$$$\mathrm{let}\:\mathrm{z}=\frac{\mathrm{z}_{\mathrm{1}} ^{\mathrm{4}} }{\mathrm{z}_{\mathrm{2}} },\:\mathrm{then}\:\mathrm{Arg}\left(\mathrm{z}\right)=\mathrm{4}\alpha−\beta \\ $$$$\mathrm{z}=\frac{\mathrm{z}_{\mathrm{1}} ^{\mathrm{4}} }{\mathrm{z}_{\mathrm{2}} }=\frac{\left(\mathrm{5}+\mathrm{i}\right)^{\mathrm{4}} }{\mathrm{239}+\mathrm{i}}=\frac{\left(\mathrm{24}+\mathrm{10i}\right)^{\mathrm{2}} \left(\mathrm{239}−\mathrm{i}\right)}{\mathrm{239}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:=\frac{\left(\mathrm{476}+\mathrm{480i}\right)\left(\mathrm{239}−\mathrm{i}\right)}{\mathrm{239}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:=\frac{\left(\mathrm{476}×\mathrm{239}+\mathrm{480}\right)+\mathrm{i}\left(\mathrm{480}×\mathrm{239}−\mathrm{476}\right)}{\mathrm{239}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{z}=\frac{\mathrm{114244}}{\mathrm{239}^{\mathrm{2}} +\mathrm{1}}\:\left(\mathrm{1}+\mathrm{i}\right)\: \\ $$$$\mathrm{z}\:\mathrm{has}\:\mathrm{argument}\:\theta=\pi/\mathrm{4}\: \\ $$$$\Rightarrow\:\mathrm{4}\alpha−\beta=\frac{\pi}{\mathrm{4}}+\mathrm{k}\pi \\ $$$$\mathrm{or}\:\:\mathrm{4tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right)=\frac{\pi}{\mathrm{4}}+\mathrm{k}\pi. \\ $$
Commented by chux last updated on 01/Jul/17
mr Ajfour I really enjoy this approach
$$\mathrm{mr}\:\mathrm{Ajfour}\:\mathrm{I}\:\mathrm{really}\:\mathrm{enjoy}\:\mathrm{this}\: \\ $$$$\mathrm{approach} \\ $$
Answered by mrW1 last updated on 30/Jun/17
α=tan^(−1) (1/5) β=tan^(−1) (1/(239)) tan 2α=((2tan α)/(1−tan^2 α))=((2×(1/5))/(1−((1/5))^2 ))=(5/(12)) tan 4α=((2tan 2α)/(1−tan^2 2α))=((2×(5/(12)))/(1−((5/(12)))^2 ))=((120)/(119)) tan (4α−β)=((tan 4α−tan β)/(1+tan 4α tan β)) =((((120)/(119))−(1/(239)))/(1+((120)/(119)) ×(1/(239)))) =((120×239−119)/(119×239+120)) =((28561)/(28561)) =1 ⇒4α−β=tan^(−1) 1=(π/4)+kπ with k∈Z ⇒ 4 tan^(−1) ((1/5))−tan^(−1) ((1/(239))) =(π/4)+kπ
$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\beta=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{239}} \\ $$$$\mathrm{tan}\:\mathrm{2}\alpha=\frac{\mathrm{2tan}\:\alpha}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha}=\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{5}}}{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} }=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\mathrm{tan}\:\mathrm{4}\alpha=\frac{\mathrm{2tan}\:\mathrm{2}\alpha}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\mathrm{2}\alpha}=\frac{\mathrm{2}×\frac{\mathrm{5}}{\mathrm{12}}}{\mathrm{1}−\left(\frac{\mathrm{5}}{\mathrm{12}}\right)^{\mathrm{2}} }=\frac{\mathrm{120}}{\mathrm{119}} \\ $$$$\mathrm{tan}\:\left(\mathrm{4}\alpha−\beta\right)=\frac{\mathrm{tan}\:\mathrm{4}\alpha−\mathrm{tan}\:\beta}{\mathrm{1}+\mathrm{tan}\:\mathrm{4}\alpha\:\mathrm{tan}\:\beta} \\ $$$$=\frac{\frac{\mathrm{120}}{\mathrm{119}}−\frac{\mathrm{1}}{\mathrm{239}}}{\mathrm{1}+\frac{\mathrm{120}}{\mathrm{119}}\:×\frac{\mathrm{1}}{\mathrm{239}}} \\ $$$$=\frac{\mathrm{120}×\mathrm{239}−\mathrm{119}}{\mathrm{119}×\mathrm{239}+\mathrm{120}} \\ $$$$=\frac{\mathrm{28561}}{\mathrm{28561}} \\ $$$$=\mathrm{1} \\ $$$$\Rightarrow\mathrm{4}\alpha−\beta=\mathrm{tan}^{−\mathrm{1}} \mathrm{1}=\frac{\pi}{\mathrm{4}}+\mathrm{k}\pi\:\mathrm{with}\:\mathrm{k}\in\mathbb{Z} \\ $$$$\Rightarrow\:\mathrm{4}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right)\:=\frac{\pi}{\mathrm{4}}+\mathrm{k}\pi \\ $$
Commented by chux last updated on 30/Jun/17
wow..... i really appreciate.
$$\mathrm{wow}…..\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$$$ \\ $$
Commented by Abbas-Nahi last updated on 01/Jul/17
very very Good.....with my best wishes
$${very}\:{very}\:\mathrm{Good}…..\mathrm{with}\:\mathrm{my}\:\mathrm{best}\:\mathrm{wishes} \\ $$

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