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Category: Trigonometry

Question-175637

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-175571

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}}…

prove-that-2tan-1-a-b-a-b-tan-2-cos-1-b-acos-a-bcos-

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}}…

tan-1-asin-sin-1-b-find-

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} \\…