Question Number 176092 by a.lgnaoui last updated on 12/Sep/22 $$\angle{AOB}\:\:\:{triangle}\:{equilareral}\:{de}\:{cote}\:\boldsymbol{{a}} \\ $$$$\left({A};\mathrm{B}\right)\::\:\mathrm{c}{e}\mathrm{ntres}\:\mathrm{de}\:\mathrm{cercles}\:\mathrm{de}\:\mathrm{rayon}\:\boldsymbol{\mathrm{r}} \\ $$$$\boldsymbol{\theta}\:=\measuredangle\:\:{GOH}\:\:\:\mathrm{EF}=\boldsymbol{{a}}_{\mathrm{0}} \\ $$$${Determiner}\:{l}\:{aire}\:{de}\:{l}\:{espace}\:{delimite}\:{par} \\ $$$$\:\:\boldsymbol{\mathrm{AFOEBHG}}\:{comme}\:{il}\:{est}\:{marque}\:{sur}\:{l}\:{image}\:{ci}−{joint}\: \\ $$$${en}\:{fonction}\:{de}:\:\boldsymbol{{a}},\boldsymbol{{r}},\boldsymbol{{a}}_{\mathrm{0}} \:{et}\:\boldsymbol{\theta} \\ $$ Commented by…
Question Number 44920 by peter frank last updated on 06/Oct/18 $$\boldsymbol{\mathrm{solve}}.\mathrm{3}^{\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} +\mathrm{3}^{\mathrm{1}−\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} =\mathrm{28} \\ $$ Answered by ajfour last updated on 06/Oct/18 $$\mathrm{3}^{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{2cos}\:^{\mathrm{2}}…
Question Number 175961 by pete last updated on 10/Sep/22 $$\mathrm{Given}\:\mathrm{that}\:\frac{\mathrm{2tan15}^{\mathrm{0}} }{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{15}^{\mathrm{0}} }=\mathrm{sin30}^{\mathrm{0}} ,\:\mathrm{find}\:\mathrm{tan15}^{\mathrm{0}} \\ $$$$\mathrm{leaving}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form}\:\left(\mathrm{radicals}\right) \\ $$ Answered by BaliramKumar last updated on 10/Sep/22…
Question Number 175956 by mnjuly1970 last updated on 10/Sep/22 Commented by mnjuly1970 last updated on 10/Sep/22 $${prove}\:\upuparrows\upuparrows \\ $$ Answered by behi834171 last updated on…
Question Number 175904 by BaliramKumar last updated on 09/Sep/22 $$ \\ $$$$\mathrm{If}\:\:\mathrm{1}\:+\:{sinx}\:+\:{sin}^{\mathrm{2}} {x}\:+\:{sin}^{\mathrm{3}} {x}\:+\:……..\:\infty\:\:\:=\:\:\mathrm{4}\:+\:\mathrm{2}\sqrt{\mathrm{3}},\: \\ $$$${x}\:=\:? \\ $$ Commented by infinityaction last updated on 09/Sep/22…
Question Number 110293 by I want to learn more last updated on 28/Aug/20 $$\mathrm{Simplify}:\:\:\:\frac{\mathrm{tan}\frac{\mathrm{3}\pi}{\mathrm{7}}\:\:\:−\:\:\mathrm{4sin}\frac{\pi}{\mathrm{7}}}{\mathrm{tan}\frac{\mathrm{6}\pi}{\mathrm{7}}\:\:+\:\:\mathrm{4sin}\frac{\mathrm{2}\pi}{\mathrm{7}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 175756 by mnjuly1970 last updated on 06/Sep/22 Commented by mr W last updated on 06/Sep/22 $${i}\:{think}\:{for}\:{any}\:{type}\:{of}\:{triangle}, \\ $$$$\frac{{r}}{{R}}\:{is}\:{maximum}\:{when}\:{A}={B}={C}=\mathrm{60}°. \\ $$ Answered by mr…
Question Number 175650 by mnjuly1970 last updated on 04/Sep/22 $$\: \\ $$$$ \\ $$$${cos}\left(\mathrm{5}{x}\right)=\:{a}.{cos}^{\:\mathrm{5}} \left({x}\right)+{b}.{cos}^{\:\mathrm{4}} \left({x}\right)+{c}.{cos}^{\mathrm{3}} \left({x}\right)+\:{d}.{cos}^{\:\mathrm{2}} \left({x}\right)+{e}.{cos}\left({x}\right)+{f} \\ $$$$\:\:\:\:\:\:{a}\:,\:{b}\:,\:{c}\:,\:{d}\:,\:{e}\:,\:{f}\:=? \\ $$$$ \\ $$$$ \\…
Question Number 175637 by mnjuly1970 last updated on 04/Sep/22 Answered by behi834171 last updated on 04/Sep/22 $$\boldsymbol{{I}}. \\ $$$$\boldsymbol{{r}}_{\boldsymbol{{a}}} =\boldsymbol{{p}}.\boldsymbol{{tg}}\frac{\boldsymbol{{A}}}{\mathrm{2}}=\boldsymbol{{p}}.\frac{\sqrt{\frac{\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\boldsymbol{{bc}}}}}{\:\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}{\boldsymbol{{bc}}}}}=\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}}= \\ $$$$=\frac{\boldsymbol{{S}}}{\boldsymbol{{p}}−\boldsymbol{{a}}}\:\:\:\:\boldsymbol{{and}}\:\boldsymbol{{so}}\:\boldsymbol{{on}}…. \\ $$$$\Rightarrow\Sigma\boldsymbol{{r}}_{\boldsymbol{{a}}} =\boldsymbol{{S}}.\Sigma\left(\frac{\mathrm{1}}{\boldsymbol{{p}}−\boldsymbol{{a}}}\right)=\boldsymbol{{S}}.\frac{\Sigma\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\Pi\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}=…
Question Number 175595 by mnjuly1970 last updated on 03/Sep/22 Terms of Service Privacy Policy Contact: info@tinkutara.com