Question Number 182045 by Emrice last updated on 03/Dec/22 $$ \\ $$$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{1}}{\mathrm{1}+\left({tanx}\right)^{\mathrm{2013}} }\:{dx}\:=\:? \\ $$$$ \\ $$ Commented by Frix last updated on…
Question Number 116513 by bounhome last updated on 04/Oct/20 $${give}\:{a}\in{Z}\:{and}\:{prove}\:{that}\:{if}\:\left({a}−\mathrm{2}\right)\mid\mathrm{4}\:{than}\:\left({a}−\mathrm{3}\right){can}'{t}\mid\mathrm{6} \\ $$ Answered by floor(10²Eta[1]) last updated on 04/Oct/20 $$\mathrm{a}−\mathrm{2}\mid\mathrm{4}\Rightarrow\mathrm{a}−\mathrm{2}\in\left\{\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{4}\right\} \\ $$$$\Rightarrow\mathrm{a}\in\mathrm{A}:\left\{−\mathrm{2},\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{4},\mathrm{6}\right\} \\ $$$$\mathrm{a}−\mathrm{3}\mid\mathrm{6}\Rightarrow\mathrm{a}−\mathrm{3}\in\left\{\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{3},\pm\mathrm{6}\right\} \\…
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Question Number 182039 by liuxinnan last updated on 03/Dec/22 $${if}\:{f}'''\left({x}\right)={f}\left({x}\right) \\ $$$${find}\:{what}\:{is}\:{f}\left({x}\right)\: \\ $$ Answered by FelipeLz last updated on 03/Dec/22 $${f}'''\left({x}\right)−{f}\left({x}\right)\:=\:\mathrm{0} \\ $$$${f}\left({x}\right)\:=\:{e}^{{rx}} \:\rightarrow\:{r}^{\mathrm{3}}…
Question Number 116486 by ZiYangLee last updated on 04/Oct/20 Answered by Dwaipayan Shikari last updated on 04/Oct/20 $$\frac{\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)}{\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)}=\frac{\mathrm{sinAcosB}+\mathrm{sinBcosA}}{\mathrm{cosAcosB}−\mathrm{sinAsinB}}=\frac{\frac{\mathrm{sinAcosB}}{\mathrm{cosAcosB}}+\frac{\mathrm{sinBcosA}}{\mathrm{cosAcosB}}}{\mathrm{1}−\mathrm{tanAtanB}} \\ $$$$=\frac{\mathrm{tanA}+\mathrm{tanB}}{\mathrm{1}−\mathrm{tanAtanB}} \\ $$ Answered by Olaf…
Question Number 182011 by mathocean1 last updated on 03/Dec/22 $${Is}\:{this}\:{true}\:{for}\:{complex}\:{number}: \\ $$$$\mathrm{4}\mathscr{R}{e}\left({z}_{\mathrm{1}} \overline {{z}_{\mathrm{2}} }\right)=\mid{z}_{\mathrm{1}} +\overline {{z}_{\mathrm{2}} }\mid^{\mathrm{2}} −\mid{z}_{\mathrm{1}} −\overline {{z}_{\mathrm{2}} }\mid^{\mathrm{2}} \\ $$ Commented…
Question Number 116459 by nguyenthanh last updated on 04/Oct/20 Answered by bemath last updated on 04/Oct/20 $$\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{12}}\right)\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\:+\left(\sqrt{\mathrm{6}}\:−\sqrt{\mathrm{2}}\:\right)\mathrm{sin}\:\left(\mathrm{x}−\frac{\mathrm{7}\pi}{\mathrm{12}}\right)\:=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{12}}\right)\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\:+\:\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)\mathrm{sin}\:\left(\mathrm{x}−\frac{\mathrm{7}\pi}{\mathrm{12}}\right)=−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{12}}\right) \\ $$$$\mathrm{1}+\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)\mathrm{cos}\:\mathrm{x}.\mathrm{sin}\:\left(\mathrm{x}−\frac{\mathrm{7}\pi}{\mathrm{12}}\right)=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{12}}\right)\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}.\mathrm{sin}\:\left(\frac{\pi}{\mathrm{12}}\right) \\ $$$$\mathrm{1}+\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}.\left(\mathrm{2cos}\:\mathrm{x}\:\mathrm{sin}\:\left(\pi−\left(\mathrm{x}−\frac{\mathrm{7}\pi}{\mathrm{12}}\right)\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right)\right. \\ $$$$\mathrm{1}+\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}.\left(\mathrm{2sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{12}}−\mathrm{x}\right).\mathrm{cos}\:\mathrm{x}\right)=\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{12}}\right)…
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Question Number 181973 by Emrice last updated on 02/Dec/22 $$ \\ $$please how can I get the proofs of the Putnam competition in pdf format?…