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Question Number 152551 by alisiao last updated on 29/Aug/21

$$\frac{sin\left(x\right)+sin\left(2x\right)+....+sin\left(nx\right)}{cos\left(x\right)+cos\left(2x\right)+....+cos\left(nx\right)}=?$$

Answered by qaz last updated on 29/Aug/21

$$\frac{\mathrm{S}}{\mathrm{C}}=\frac{\mathrm{\Im }({\mathrm{e}}^{\mathrm{ix}}+{\mathrm{e}}^{2\mathrm{ix}}+...+{\mathrm{e}}^{\mathrm{nix}})}{\mathrm{\Re }({\mathrm{e}}^{\mathrm{ix}}+{\mathrm{e}}^{2\mathrm{ix}}+...+{\mathrm{e}}^{\mathrm{nix}})}$$$$=\frac{\mathrm{\Im }\frac{{\mathrm{e}}^{\mathrm{ix}}(1-{\mathrm{e}}^{\mathrm{nix}})}{1-{\mathrm{e}}^{\mathrm{ix}}}}{\mathrm{\Re }\frac{{\mathrm{e}}^{\mathrm{ix}}(1-{\mathrm{e}}^{\mathrm{nix}})}{1-{\mathrm{e}}^{\mathrm{ix}}}}$$$$=\frac{\mathrm{\Im }\frac{{\mathrm{e}}^{\mathrm{ix}}(1-{\mathrm{e}}^{\mathrm{nix}})(1-{\mathrm{e}}^{-\mathrm{ix}})}{(1-{\mathrm{e}}^{\mathrm{ix}})(1-{\mathrm{e}}^{-\mathrm{ix}})}}{\mathrm{\Re }\frac{{\mathrm{e}}^{\mathrm{ix}}(1-{\mathrm{e}}^{\mathrm{nix}})(1-{\mathrm{e}}^{-\mathrm{ix}})}{(1-{\mathrm{e}}^{\mathrm{ix}})(1-{\mathrm{e}}^{-\mathrm{ix}})}}$$$$=\frac{\mathrm{\Im }({\mathrm{e}}^{\mathrm{ix}}-1)({\mathrm{e}}^{\mathrm{nix}}-1)}{\mathrm{\Re }({\mathrm{e}}^{\mathrm{ix}}-1)({\mathrm{e}}^{\mathrm{nix}}-1)}$$$$=\frac{\mathrm{\Im }({\mathrm{e}}^{(\mathrm{n}+1)\mathrm{ix}}-{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{\mathrm{nix}}+1)}{\mathrm{\Re }({\mathrm{e}}^{(\mathrm{n}+1)\mathrm{ix}}-{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{\mathrm{nix}}+1)}$$$$=\frac{\sin (\mathrm{n}+1)\mathrm{x}-\sin \mathrm{x}-\sin \mathrm{nx}}{\cos (\mathrm{n}+1)\mathrm{x}-\cos \mathrm{x}-\cos \mathrm{nx}+1}$$

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