Question Number 178779 by Laftex234 last updated on 21/Oct/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 178743 by mathlove last updated on 21/Oct/22 $${f}\left({x}\right)={arctan}\left(\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}}\right)\:\: \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$ Commented by cortano1 last updated on 21/Oct/22 $$\:\mathrm{tan}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{2x}+\mathrm{1}}} \\ $$$$\:\mathrm{tan}\:^{\mathrm{2}}…
Question Number 113200 by bobhans last updated on 11/Sep/20 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{2tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)=\frac{\pi}{\mathrm{4}} \\ $$ Answered by john santu last updated on 11/Sep/20 $$\left({Q}\right)\:{prove}\:{that}\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)\:=\:\frac{\pi}{\mathrm{4}}.…
Question Number 178694 by peter frank last updated on 20/Oct/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\mathrm{2}^{\mathrm{cos}\:\mathrm{2}\theta} +\mathrm{1}=\mathrm{3}.\mathrm{2}^{−\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$ Answered by Frix last updated on 20/Oct/22 $$\mathrm{2}^{\mathrm{cos}\:\mathrm{2}\theta}…
Question Number 47586 by Rio Michael last updated on 11/Nov/18 $${Given}\:{the}\:{function}\:{f}\left(\theta\right)=\:{cos}\mathrm{2}\theta\:−\:{sin}\theta.\:\left({for}\:\mathrm{0}°\leqslant\theta\leqslant\pi\right) \\ $$$${plot}\:{the}\:{graph}\:{for}\:\:{intervals}\:{of}\:\frac{\pi}{\mathrm{6}}. \\ $$$${hence}\:{find}\:{the}\:{value}\:{of}\:{cos}\mathrm{2}\theta={sin}\theta. \\ $$ Answered by Joel578 last updated on 12/Nov/18 Commented…
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}}…