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Question-119866




Question Number 119866 by Dwaipayan Shikari last updated on 27/Oct/20
Commented by Ar Brandon last updated on 27/Oct/20
Where did you get this, bro ?����
Commented by Dwaipayan Shikari last updated on 27/Oct/20
On Brilliant app
$${On}\:{Brilliant}\:{app} \\ $$
Commented by Dwaipayan Shikari last updated on 27/Oct/20
https://brilliant.org/problems/functional-complex-trignometry
Commented by Ar Brandon last updated on 27/Oct/20
��OK cool
Answered by TANMAY PANACEA last updated on 27/Oct/20
tan(θ_1 +θ_2 +...+θ_(52) )=((s_1 −s_3 +s_5 −s_7 +s_9 ...+(−1)^(26−1) s_(51) )/(1−s_2 +s_4 −s_6 +..+(−1)^(26) s_(52) ))  51=1+(n−1)2→n=26  52=2+(k−1)2→k=26  s_1 =−a_1   s_2 =a_2   s_3 =−a_3   s_4 =a_4   tan(θ_1 +θ_2 +..+θ_(52) )  =((−a_1 +a_3 −a_5 ...)/(1−a_2 +a_4 −..))  f(x)=x^(52) +a_1 x^(51) +a_2 x^(50) +a_3 x^(49) +..+a_(52)   put x=1  f(1)=1+a_1 +a_2 +a_3 +..+a_(52)   f(i)=(i)^(52) +a_1 (i)^(51) +a_2 (i)^(50) +a_3 (i)^(49) +..  i(2+(√3) )−1=1+a_1 ×(i^2 )^(25) ×i+a_2 ×(i^2 )^(25) +a_3 ×(i^2 )^(24) (i)+..  wait...  i have to solve in register...
$${tan}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} +…+\theta_{\mathrm{52}} \right)=\frac{{s}_{\mathrm{1}} −{s}_{\mathrm{3}} +{s}_{\mathrm{5}} −{s}_{\mathrm{7}} +{s}_{\mathrm{9}} …+\left(−\mathrm{1}\right)^{\mathrm{26}−\mathrm{1}} {s}_{\mathrm{51}} }{\mathrm{1}−{s}_{\mathrm{2}} +{s}_{\mathrm{4}} −{s}_{\mathrm{6}} +..+\left(−\mathrm{1}\right)^{\mathrm{26}} {s}_{\mathrm{52}} } \\ $$$$\mathrm{51}=\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{2}\rightarrow{n}=\mathrm{26} \\ $$$$\mathrm{52}=\mathrm{2}+\left({k}−\mathrm{1}\right)\mathrm{2}\rightarrow{k}=\mathrm{26} \\ $$$${s}_{\mathrm{1}} =−{a}_{\mathrm{1}} \:\:{s}_{\mathrm{2}} ={a}_{\mathrm{2}} \:\:{s}_{\mathrm{3}} =−{a}_{\mathrm{3}} \:\:{s}_{\mathrm{4}} ={a}_{\mathrm{4}} \\ $$$${tan}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} +..+\theta_{\mathrm{52}} \right) \\ $$$$=\frac{−{a}_{\mathrm{1}} +{a}_{\mathrm{3}} −{a}_{\mathrm{5}} …}{\mathrm{1}−{a}_{\mathrm{2}} +{a}_{\mathrm{4}} −..} \\ $$$${f}\left({x}\right)={x}^{\mathrm{52}} +{a}_{\mathrm{1}} {x}^{\mathrm{51}} +{a}_{\mathrm{2}} {x}^{\mathrm{50}} +{a}_{\mathrm{3}} {x}^{\mathrm{49}} +..+{a}_{\mathrm{52}} \\ $$$${put}\:{x}=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +..+{a}_{\mathrm{52}} \\ $$$${f}\left({i}\right)=\left({i}\right)^{\mathrm{52}} +{a}_{\mathrm{1}} \left({i}\right)^{\mathrm{51}} +{a}_{\mathrm{2}} \left({i}\right)^{\mathrm{50}} +{a}_{\mathrm{3}} \left({i}\right)^{\mathrm{49}} +.. \\ $$$${i}\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)−\mathrm{1}=\mathrm{1}+{a}_{\mathrm{1}} ×\left({i}^{\mathrm{2}} \right)^{\mathrm{25}} ×{i}+{a}_{\mathrm{2}} ×\left({i}^{\mathrm{2}} \right)^{\mathrm{25}} +{a}_{\mathrm{3}} ×\left({i}^{\mathrm{2}} \right)^{\mathrm{24}} \left({i}\right)+.. \\ $$$$\boldsymbol{{wait}}… \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{have}}\:\boldsymbol{{to}}\:\boldsymbol{{solve}}\:\boldsymbol{{in}}\:\boldsymbol{{register}}… \\ $$$$ \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 27/Oct/20
I think (tanθ_1 +tanθ_2 +...)_(min) =(θ_1 +θ_2 +θ_3 +...)_(min)   Small angle . Not sure
$${I}\:{think}\:\left({tan}\theta_{\mathrm{1}} +{tan}\theta_{\mathrm{2}} +…\right)_{{min}} =\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} +\theta_{\mathrm{3}} +…\right)_{{min}} \\ $$$${Small}\:{angle}\:.\:{Not}\:{sure}\: \\ $$

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