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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

$$OnBrilliantapp$$

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

$$tan({\theta }_1+{\theta }_2+...+{\theta }_{52})=\frac{s_1-s_3+s_5-s_7+s_9...+{(-1)}^{26-1}{s}_{51}}{1-s_2+s_4-s_6+..+{(-1)}^{26}{s}_{52}}$$$$51=1+(n-1)2\to n=26$$$$52=2+(k-1)2\to k=26$$$$s_1=-a_1{s}_2=a_2{s}_3=-a_3{s}_4=a_4$$$$tan({\theta }_1+{\theta }_2+..+{\theta }_{52})$$$$=\frac{-a_1+a_3-a_5...}{1-a_2+a_4-..}$$$$f\left(x\right)=x^{52}+a_1{x}^{51}+a_2{x}^{50}+a_3{x}^{49}+..+a_{52}$$$$putx=1$$$$f\left(1\right)=1+a_1+a_2+a_3+..+a_{52}$$$$f\left(i\right)={\left(i\right)}^{52}+a_1{\left(i\right)}^{51}+a_2{\left(i\right)}^{50}+a_3{\left(i\right)}^{49}+..$$$$i(2+\sqrt{3})-1=1+a_1\times {\left(i^2\right)}^{25}\times i+a_2\times {\left(i^2\right)}^{25}+a_3\times {\left(i^2\right)}^{24}\left(i\right)+..$$$$\mathit{w}\mathit{a}\mathit{i}\mathit{t}...$$$$\mathit{i}\mathit{h}\mathit{a}\mathit{v}\mathit{e}\mathit{t}\mathit{o}\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}\mathit{i}\mathit{n}\mathit{r}\mathit{e}\mathit{g}\mathit{i}\mathit{s}\mathit{t}\mathit{e}\mathit{r}...$$$$$$$$$$

Commented by Dwaipayan Shikari last updated on 27/Oct/20

$$Ithink{(tan{\theta }_1+tan{\theta }_2+...)}_{min}={({\theta }_1+{\theta }_2+{\theta }_3+...)}_{min}$$$$Smallangle.Notsure$$

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