Question Number 112885 by malwan last updated on 10/Sep/20 Commented by malwan last updated on 10/Sep/20 $${P}=? \\ $$$$\left({a}\right)\:\left(−\mathrm{5}\pi/\mathrm{4},\mathrm{2}^{\mathrm{1}/\mathrm{2}} \right) \\ $$$$\left({b}\right)\:\left(−\mathrm{5}\pi/\mathrm{4},\mathrm{2}^{−\mathrm{1}/\mathrm{2}} \right) \\ $$$$\left({c}\right)\:\left(−\mathrm{7}\pi/\mathrm{4},\mathrm{2}^{\mathrm{1}/\mathrm{2}}…
Question Number 178388 by mathlove last updated on 16/Oct/22 $$\:{p}\left({sin}^{\mathrm{77}} \frac{\mathrm{50}}{\mathrm{19}},{cos}\frac{\mathrm{27}}{\mathrm{13}}\right) \\ $$$$\left.\mathrm{1}\left.\right)\left.\:\left.{VI}\:\:\:\:\mathrm{2}\right){III}\:\:\:\:\:\mathrm{3}\right){II}\:\:\:\:\:\:\:\mathrm{4}\right){I} \\ $$In which region of the fixed coordinate system is this point…
Question Number 47150 by somil last updated on 05/Nov/18 Commented by maxmathsup by imad last updated on 05/Nov/18 $${we}\:{see}\:{that}\:{cos}\theta\:{must}\:{be}\:\geqslant\mathrm{0}\Rightarrow\theta\:\in\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right]\left[\mathrm{2}\pi\right]\:\:{changement}\:={x}\:={sin}\theta\:{give} \\ $$$$\mathrm{3}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{2}{x}^{\mathrm{2}} \:\Rightarrow\mathrm{9}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\mathrm{4}{x}^{\mathrm{4}} \:\Rightarrow\mathrm{4}{x}^{\mathrm{4}}…
Question Number 112672 by bemath last updated on 09/Sep/20 $$\:\frac{\mathrm{sin}\:\mathrm{A}−\mathrm{cos}\:\mathrm{A}+\mathrm{1}}{\mathrm{sin}\:\mathrm{A}+\mathrm{cos}\:\mathrm{A}−\mathrm{1}}\:?\: \\ $$ Commented by som(math1967) last updated on 09/Sep/20 $$\frac{\mathrm{1}+\mathrm{sinA}}{\mathrm{cosA}} \\ $$ Commented by bemath…
Question Number 178181 by mathlove last updated on 13/Oct/22 $$\frac{{cos}\alpha\:{cot}\alpha−{sin}\alpha\:{tan}\alpha}{{csc}\alpha\:{sec}\alpha}=? \\ $$ Commented by Adedayo2000 last updated on 13/Oct/22 $${Recall}\:{cot}\alpha=\frac{\mathrm{1}}{{tan}\alpha}=\frac{{cos}\alpha}{{sin}\alpha},\:{csc}\alpha=\frac{\mathrm{1}}{{sin}\alpha},\:{sec}\alpha=\frac{\mathrm{1}}{{cos}\alpha} \\ $$$$=>\frac{\left({cos}\alpha\right)\left(\frac{{cos}\alpha}{{sin}\alpha}\right)−\left({sin}\alpha\right)\left(\frac{{sin}\alpha}{{cos}\alpha}\right)}{\left(\frac{\mathrm{1}}{{sin}\alpha}\right)\left(\frac{\mathrm{1}}{{cos}\alpha}\right)} \\ $$$$=>\frac{\frac{{cos}^{\mathrm{2}} \alpha}{{sin}\alpha}−\frac{{sin}^{\mathrm{2}}…
Question Number 112645 by bemath last updated on 09/Sep/20 $$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\:=?\: \\ $$ Answered by john santu last updated on 09/Sep/20 $${let}\:{p}\:=\:\frac{\pi}{\mathrm{7}}\:{and}\:{X}=\mathrm{cos}\:{p}−\mathrm{cos}\:\mathrm{2}{p}+\mathrm{cos}\:\mathrm{3}{p} \\ $$$${multiply}\:{by}\:\mathrm{2sin}\:{p} \\ $$$$\mathrm{2sin}\:{p}\:{X}\:=\:\mathrm{2sin}\:{p}\left(\mathrm{cos}\:{p}−\mathrm{cos}\:\mathrm{2}{p}+\mathrm{cos}\:\mathrm{3}{p}\right)…
Question Number 178182 by mathlove last updated on 13/Oct/22 $$\frac{\mathrm{1}}{{cos}\mathrm{80}}−\frac{\sqrt{\mathrm{3}}}{{sin}\mathrm{80}}=? \\ $$ Commented by cortano1 last updated on 13/Oct/22 $$\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{10}°}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{cos}\:\:\mathrm{10}°}\:=\:\frac{\mathrm{cos}\:\:\mathrm{10}°−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{10}}{\mathrm{sin}\:\mathrm{10}°\:\mathrm{cos}\:\mathrm{10}°} \\ $$$$\:=\:\frac{\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{10}°−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:=\:\frac{\mathrm{4}\left(\mathrm{sin}\:\mathrm{30}°\:\mathrm{cos}\:\mathrm{10}°−\mathrm{cos}\:\mathrm{30}°\:\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\…
Question Number 178153 by mnjuly1970 last updated on 13/Oct/22 Answered by mr W last updated on 13/Oct/22 $$\mathrm{2}=\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\geqslant\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow{a}\geqslant\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}\:\Rightarrow{a}\geqslant\mathrm{4} \\ $$$$\mathrm{2}=\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\leqslant\left(\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow{c}\leqslant\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}\:\Rightarrow{c}\leqslant\mathrm{3} \\…
Question Number 178091 by alcohol last updated on 12/Oct/22 $${remainder}\:{when}\:\mathrm{7}^{\mathrm{9}^{\mathrm{9}^{\mathrm{9}^{\mathrm{9}} } } } \:{is}\:{divided}\:{by}\:\mathrm{10}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 178087 by a.lgnaoui last updated on 12/Oct/22 $${un}\:{avion}\:{volant}\:{a}\:{hauteur}\:{h}=\mathrm{4}{km}\:{du}\:{sol} \\ $$$${a}\:{vitesse}\:{supersonic}\:\:\left({v}\right)\: \\ $$$${sachant}\:{que}\:{la}\:{vitesse}\:{du}\:{son}\:{a}\:{la}\: \\ $$$${hauteur}\:{h}\:{est}\:{de}\:\mathrm{300}{m}/{s}\:{et} \\ $$$${que}\:{le}\:{temps}\:{mis}\:{par}\:{le}\:{son}\:{emis}\:{par} \\ $$$${l}\:{avion}\:{depuis}\:{un}\:{point}\left({c}\right)\:{est}\: \\ $$$${cspte}\:{par}\:{une}\:{prrsonne}\:{avec}\:{un} \\ $$$${retard}\:{de}\:\mathrm{3}{s} \\…