Question Number 116023 by aye48 last updated on 30/Sep/20 $$\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:+\:\frac{\mathrm{x}}{\mathrm{a}}\:\left(\mathrm{1}\:+\:\mathrm{x}\right)+\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:\left(\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \right)+\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{a}^{\mathrm{3}} }\:\left(\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \right)\:+\:\ldots \\ $$$$ \\ $$ Answered by…
Question Number 50397 by Abdo msup. last updated on 16/Dec/18 $${calculate}\:{artan}\left(\mathrm{2}\right)+{arctan}\left(\mathrm{5}\right)+{arctan}\left(\mathrm{8}\right) \\ $$ Answered by mr W last updated on 17/Dec/18 $${x}=\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{tan}^{−\mathrm{1}} \mathrm{8} \\…
Question Number 115891 by bemath last updated on 29/Sep/20 $$\mathrm{sec}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{cosec}\:^{\mathrm{2}} \mathrm{20}°+\mathrm{cosec}\:^{\mathrm{2}} \mathrm{40}°−\mathrm{sec}\:^{\mathrm{2}} \mathrm{45}° \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115798 by aye48 last updated on 28/Sep/20 Commented by bemath last updated on 29/Sep/20 $$\mathrm{sin}\:\alpha.\mathrm{cos}\:\beta\:=\:\frac{\mathrm{9}}{\mathrm{10}} \\ $$$$\Rightarrow\mathrm{2sin}\:\alpha.\mathrm{cos}\:\beta\:=\:\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\alpha+\beta\right)+\mathrm{sin}\:\left(\alpha−\beta\right)\:=\:\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\alpha+\beta\right)\:=\:\frac{\mathrm{9}}{\mathrm{5}}\:−\:\frac{\mathrm{4}}{\mathrm{5}}\:=\:\mathrm{1} \\ $$…
Question Number 115765 by aye48 last updated on 28/Sep/20 $$\: \\ $$$$\:\:\:\mathrm{Show}\:\mathrm{that}\:\mathrm{sin}\:\left(\alpha\:+\:\beta\right)\:=\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\:+\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta. \\ $$ Commented by bemath last updated on 28/Sep/20 Commented by aye48 last…
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}…