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}}. \\ $$$$\boldsymbol{{r}}_{\boldsymbol{{a}}} =\boldsymbol{{p}}.\boldsymbol{{tg}}\frac{\boldsymbol{{A}}}{\mathrm{2}}=\boldsymbol{{p}}.\frac{\sqrt{\frac{\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\boldsymbol{{bc}}}}}{\:\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}{\boldsymbol{{bc}}}}}=\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}}= \\ $$$$=\frac{\boldsymbol{{S}}}{\boldsymbol{{p}}−\boldsymbol{{a}}}\:\:\:\:\boldsymbol{{and}}\:\boldsymbol{{so}}\:\boldsymbol{{on}}…. \\ $$$$\Rightarrow\Sigma\boldsymbol{{r}}_{\boldsymbol{{a}}} =\boldsymbol{{S}}.\Sigma\left(\frac{\mathrm{1}}{\boldsymbol{{p}}−\boldsymbol{{a}}}\right)=\boldsymbol{{S}}.\frac{\Sigma\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\Pi\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}=…
Question Number 175595 by mnjuly1970 last updated on 03/Sep/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 175585 by mnjuly1970 last updated on 03/Sep/22 $$ \\ $$$$\:\:\:{in}\:{A}\overset{\Delta} {{B}C}\:\:{prove}\:\:{that}: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:{sin}\:\left(\frac{\:{A}}{\mathrm{2}}\:\right)\:\leqslant\:\frac{\:{a}}{\:{b}\:+\:{c}}\:\:\:\:\:\:\:\blacksquare \\ $$$$ \\ $$ Commented by mahdipoor last…
Question Number 175571 by mnjuly1970 last updated on 02/Sep/22 Answered by behi834171 last updated on 03/Sep/22 $$\boldsymbol{{cos}}\frac{\boldsymbol{{A}}}{\mathrm{2}}=\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}{\boldsymbol{{bc}}}}\:\:\boldsymbol{{and}}\:\boldsymbol{{so}}\:\boldsymbol{{on}}… \\ $$$$\frac{\boldsymbol{{a}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)}{\boldsymbol{{bccos}}^{\mathrm{2}} \frac{\boldsymbol{{A}}}{\mathrm{2}}}=\frac{\boldsymbol{{a}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)}{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}\:\:\:\boldsymbol{{and}}\:\boldsymbol{{so}}\:\boldsymbol{{on}}…. \\ $$$$\Rightarrow{lhs}=\frac{\boldsymbol{{a}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)}{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}+\frac{\boldsymbol{{b}}\left(\boldsymbol{{c}}+\boldsymbol{{a}}\right)}{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)}+\frac{\boldsymbol{{c}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)}{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}= \\ $$$$\geqslant\sqrt[{\mathrm{3}}]{\frac{\boldsymbol{{abc}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)\left(\boldsymbol{{c}}+\boldsymbol{{a}}\right)}{\boldsymbol{{p}}^{\mathrm{2}} \boldsymbol{{S}}^{\mathrm{2}}…
Question Number 44478 by peter frank last updated on 29/Sep/18 $${prove}\:{that}\:\mathrm{2tan}^{−\mathrm{1}} \left(\sqrt{\frac{{a}−{b}}{{a}+{b}\:}}\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right)=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{b}+{acos}\theta}{{a}+{bcos}\theta}\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18 $${RHS} \\ $$$${cos}^{−\mathrm{1}}…
Question Number 175544 by cortano1 last updated on 02/Sep/22 $$\:\mathrm{tan}\:^{\mathrm{6}} \left(\mathrm{10}°\right)+\mathrm{tan}\:^{\mathrm{6}} \left(\mathrm{50}°\right)+\mathrm{tan}\:^{\mathrm{6}} \left(\mathrm{70}°\right)=? \\ $$ Commented by Frix last updated on 29/Jan/23 $$\mathrm{433} \\ $$…
Question Number 175493 by ajfour last updated on 31/Aug/22 $$\mathrm{tan}^{−\mathrm{1}} \left({a}\mathrm{sin}\:\theta\right)=\mathrm{sin}^{−\mathrm{1}} {b}−\theta \\ $$$${find}\:\theta. \\ $$ Commented by Frix last updated on 01/Sep/22 $$\mathrm{I}\:\mathrm{tried}\:\mathrm{but}\:\mathrm{it}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{4th}\:\mathrm{degree}\:\mathrm{polynome} \\…