Question Number 49963 by maxmathsup by imad last updated on 12/Dec/18 $${find}\:\:{cos}\left(\frac{\pi}{\mathrm{7}}\right)\:{by}\:{solving}\:{the}\:{equation}\:{x}^{\mathrm{7}} −\mathrm{1}=\mathrm{0}\:{inside}\:{C}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115455 by bemath last updated on 26/Sep/20 $$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{32}\pi}{\mathrm{65}}\right)=? \\ $$ Commented by Adel last updated on 13/Jan/21 $$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{32}\pi}{\mathrm{65}}\right)=? \\ $$ Answered by TANMAY…
Question Number 115348 by bemath last updated on 25/Sep/20 $${If}\:{x}\:\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right)\:{and}\:\mathrm{2cos}\:{x}\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)+\mathrm{tan}\:^{\mathrm{2}} {x}\:<\:\mathrm{sec}\:^{\mathrm{2}} {x}\: \\ $$$${has}\:{solution}\:{set}\:{is}\:{a}<{x}<{b}.\:{find}\:{the} \\ $$$${value}\:{of}\:{a}+{b} \\ $$ Answered by bobhans last updated on 25/Sep/20…
Question Number 115345 by bemath last updated on 25/Sep/20 $$\mathrm{sec}\:\theta\:\left(\mathrm{sec}\:\theta\:\left(\mathrm{sin}\:^{\mathrm{2}} \theta\right)+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)=\mathrm{1} \\ $$$${has}\:{the}\:{roots}\:{are}\:\theta_{\mathrm{1}} \:{and}\:\theta_{\mathrm{2}} .\:{Find}\:{the} \\ $$$${value}\:{of}\:\mathrm{tan}\:\theta_{\mathrm{1}} ×\mathrm{tan}\:\theta_{\mathrm{2}} . \\ $$ Answered by bobhans last…
Question Number 115332 by bobhans last updated on 25/Sep/20 $${Minimum}\:{value}\:{of}\:{function}\: \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{16}{x}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{4}}{{x}\:\mathrm{cos}\:{x}}\:{where}\:−\pi<{x}<\mathrm{0} \\ $$ Commented by bemath last updated on 25/Sep/20 $$\Leftrightarrow\:{f}\left({x}\right)=\mathrm{16}{x}\:\mathrm{cos}\:{x}\:+\:\mathrm{4}{x}^{−\mathrm{1}} \:\mathrm{sec}\:{x}…
Question Number 115328 by bobhans last updated on 25/Sep/20 $${If}\:\frac{\mathrm{sin}\:\mathrm{1}°+\mathrm{sin}\:\mathrm{2}°+\mathrm{sin}\:\mathrm{3}°+…+\mathrm{sin}\:\mathrm{44}°}{\mathrm{cos}\:\mathrm{1}°+\mathrm{cos}\:\mathrm{2}°+\mathrm{cos}\:\mathrm{3}°+…+\mathrm{cos}\:\mathrm{44}°}=\chi \\ $$$${then}\:\chi^{\mathrm{4}} +\mathrm{4}\chi^{\mathrm{3}} +\mathrm{4}\chi^{\mathrm{2}} +\mathrm{4}= \\ $$ Answered by bemath last updated on 25/Sep/20 $$\:\:\mathrm{sin}\:\mathrm{44}°+\mathrm{sin}\:\mathrm{1}°=\mathrm{2sin}\:\left(\frac{\mathrm{45}°}{\mathrm{2}}\right).\mathrm{cos}\:\left(\frac{\mathrm{43}°}{\mathrm{2}}\right)…
Question Number 180826 by mnjuly1970 last updated on 17/Nov/22 $$ \\ $$$$\:\:\:\:\mathrm{I}{f}\:\:,\:\:\:\mathrm{2}{sin}\left(\theta\:\right)−\mathrm{3}{cos}\left(\theta\right)\:=\mathrm{3} \\ $$$$\:\Rightarrow\:\:\:\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}\right)\:−\:\mathrm{3}{cos}\left(\frac{\theta}{\mathrm{2}}\right)\:=\:? \\ $$$$ \\ $$ Answered by mr W last updated on…
Question Number 115255 by aye48 last updated on 24/Sep/20 $$\mathrm{Express}\:\mathrm{cosec}\:\mathrm{3x}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{cosec}\:\mathrm{x}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115253 by aye48 last updated on 24/Sep/20 $$\mathrm{Express}\:\mathrm{sin}\:\mathrm{4x}\:\mathrm{interm}\:\mathrm{of}\:\mathrm{sin}\:\mathrm{x}. \\ $$ Answered by Dwaipayan Shikari last updated on 24/Sep/20 $$\mathrm{sin4x}=\mathrm{2sin2xcos2x} \\ $$$$=\mathrm{2sinxcosx}\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\right) \\…
Question Number 115248 by aye48 last updated on 24/Sep/20 $$\mathrm{Express}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{interm}\:\mathrm{of}\:\mathrm{cos}\:\mathrm{x}. \\ $$ Answered by Olaf last updated on 24/Sep/20 $$\mathrm{cos4}{x}\:=\:\mathrm{2cos}^{\mathrm{2}} \mathrm{2}{x}−\mathrm{1} \\ $$$$\mathrm{cos4}{x}\:=\:\mathrm{2}\left(\mathrm{2cos}^{\mathrm{2}} {x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}…