Question Number 86476 by DuDono last updated on 28/Mar/20 $$\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \alpha=−{i}\:\mathrm{ln}\:\left(\alpha\pm\sqrt{\alpha^{\mathrm{2}} −\mathrm{1}}\right)−\frac{\pi}{\mathrm{2}} \\ $$ Commented by MJS last updated on 28/Mar/20 $$\mathrm{the}\:\mathrm{path}\:\mathrm{is}\:\mathrm{this}: \\…
Question Number 152010 by RB95 last updated on 25/Aug/21 Commented by RB95 last updated on 25/Aug/21 $$ \\ $$$${Slt} \\ $$$${Pouviez}\:{vous}\:{m}'{aider}? \\ $$ Answered by…
Question Number 151959 by john_santu last updated on 24/Aug/21 Answered by iloveisrael last updated on 24/Aug/21 $$\left(\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} +\left(\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{1}\right)\left(\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}−\left(\mathrm{sin}\:{x}\mathrm{cos}\:{x}\right)^{\mathrm{2}}…
Question Number 20886 by tammi last updated on 06/Sep/17 $${if}\:\mathrm{sin}\:{x}={m}\mathrm{sin}\:{y} \\ $$$${so}\:{proof}\:{that} \\ $$$$\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)=\frac{{m}−\mathrm{1}}{{m}+\mathrm{1}}\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}\right) \\ $$ Answered by ajfour last updated on 06/Sep/17 $${m}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{y}}\:\:\:\:\Rightarrow\:\:\frac{{m}−\mathrm{1}}{{m}+\mathrm{1}}=\frac{\mathrm{sin}\:{x}−\mathrm{sin}\:{y}}{\mathrm{sin}\:{x}+\mathrm{sin}\:{y}} \\…
Question Number 20885 by tammi last updated on 06/Sep/17 $${if}\:\left(\theta−\varphi\right){subtle}\:{and}\:\:\:\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi= \\ $$$$\left(\mathrm{cos}\:\varphi−\mathrm{cos}\:\theta\right)\sqrt{\mathrm{3}} \\ $$$${so}\:{proof}\:\mathrm{sin}\:\mathrm{3}\theta+\mathrm{sin}\:\mathrm{3}\varphi=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20884 by tammi last updated on 06/Sep/17 $$\mathrm{2cos}\:\frac{\pi}{\mathrm{3}}\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{13}}=\mathrm{0} \\ $$ Answered by ajfour last updated on 06/Sep/17 $${must}\:{start}\:{with}\:\:\:\:\mathrm{2cos}\:\frac{\pi}{\mathrm{13}}…,\:{Then} \\ $$$$\:{L}.{H}.{S}.\:=\:\mathrm{cos}\:\left(\frac{\mathrm{10}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{13}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{13}}\right) \\…
Question Number 151941 by john_santu last updated on 24/Aug/21 $$\mathrm{Find}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{equation} \\ $$$$\:\mathrm{6cos}\:\mathrm{x}−\mathrm{8sin}\:\mathrm{x}=\mathrm{5}\sqrt{\mathrm{3}} \\ $$$$\:\mathrm{0}°\leqslant\mathrm{x}\leqslant\mathrm{360}° \\ $$ Answered by iloveisrael last updated on 24/Aug/21 $$\:\Leftrightarrow\:\mathrm{3cos}\:{x}−\mathrm{4sin}\:{x}\:=\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 151922 by john_santu last updated on 24/Aug/21 $$\:\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}}−\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\frac{\pi}{\mathrm{9}}}\:=? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 86346 by Power last updated on 28/Mar/20 Commented by MJS last updated on 28/Mar/20 $$−\mathrm{4} \\ $$ Commented by jagoll last updated on…
Question Number 20785 by Tinkutara last updated on 02/Sep/17 $$\mathrm{If}\:\mathrm{in}\:\mathrm{a}\:\Delta{ABC},\:\mathrm{tan}\frac{{A}}{\mathrm{2}},\:\mathrm{tan}\frac{{B}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{tan}\frac{{C}}{\mathrm{2}} \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{HP},\:\mathrm{then}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cot}\frac{{B}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$ Answered…