Question Number 107555 by anonymous last updated on 11/Aug/20 Answered by bshahid010 last updated on 11/Aug/20 $${if}\:{cos}\alpha=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow{sin}\alpha=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${and}\:{sin}\beta=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow{cos}\beta=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${Now}\:{cos}\left(\alpha+\beta\right)={cos}\alpha.{cos}\beta−{sin}\alpha.{sin}\beta \\ $$$$\Rightarrow−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow−\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}\right)…
Question Number 107534 by bemath last updated on 11/Aug/20 $$\:\:\:\vDash\mathcal{B}{e}\mathcal{M}{ath}\vDash \\ $$$$\:\mathrm{sin}\:^{\mathrm{6}} {x}\:+\:\mathrm{cos}\:^{\mathrm{6}} {x}\:=\:\frac{\mathrm{7}}{\mathrm{16}}\:;\:{x}\:\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$ Commented by hgrocks last updated on 11/Aug/20 7 and 16 both my favorite numbers ���� Commented…
Question Number 107470 by bemath last updated on 11/Aug/20 $$\:\:\:\:\:\:\:\:\circlearrowleft\mathcal{B}{e}\mathcal{M}{ath}\circlearrowright \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{3}°\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{4}°\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{41}°\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{42}°\right)=? \\ $$$$\left(\mathrm{2}\right){f}\left({x}\right)={g}\left({h}\left({x}\right)\right);\:{h}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}. \\ $$$${If}\:{f}\:'\left(−\mathrm{1}\right)=\mathrm{14}\:{then}\:{g}\:'\left(\mathrm{5}\right)=? \\ $$ Answered by john santu last updated…
Question Number 41902 by upadhyayrakhi20@gmail.com last updated on 15/Aug/18 $$\mathrm{tan3}\theta\:\mathrm{tan2}\theta\:=\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:^{\mathrm{solution}……} \\ $$ Answered by MJS last updated on 15/Aug/18 $$\mathrm{tan}\:\mathrm{3}\theta\:\mathrm{tan}\:\mathrm{2}\theta\:−\mathrm{1}=\mathrm{0} \\ $$$$−\mathrm{2}\frac{\mathrm{cos}\:\mathrm{5}\theta}{\mathrm{cos}\:\mathrm{5}\theta\:+\mathrm{cos}\:\theta}=\mathrm{0} \\…
Question Number 172923 by bagjagugum123 last updated on 03/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107364 by bemath last updated on 10/Aug/20 $$\:\:\:\:\:\:\circledcirc\mathscr{BEMATH}\circledcirc\: \\ $$$${If}\:\mathrm{tan}\:{x}+\mathrm{sec}\:{x}\:=\:{b}\:\Rightarrow\:\mathrm{cos}\:{x}\:=\:? \\ $$ Answered by bobhans last updated on 10/Aug/20 $$\:\:\:\:\:\:\Rrightarrow\mathbb{BOB}\mathcal{HANS}\Lleftarrow \\ $$$$\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{sec}\:\mathrm{x}}{\mathrm{tan}\:\mathrm{x}−\mathrm{sec}\:\mathrm{x}}\:×\:\mathrm{tan}\:\mathrm{x}+\mathrm{sec}\:\mathrm{x}\:=\:\flat \\…
Question Number 41691 by avishek last updated on 11/Aug/18 $${tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{33} \\ $$ Answered by ajfour last updated on 12/Aug/18 $${let}\:\mathrm{tan}\:\mathrm{20}°={m} \\ $$$$\mathrm{tan}\:\mathrm{3}\theta=\frac{\mathrm{3tan}\:\theta−\mathrm{tan}\:^{\mathrm{3}}…
Question Number 172732 by depressiveshrek last updated on 30/Jun/22 $$\mathrm{sin}\theta−\mathrm{cos}\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{4}} \theta+\mathrm{cos}^{\mathrm{4}} \theta=? \\ $$ Answered by floor(10²Eta[1]) last updated on 30/Jun/22 $$\left(\mathrm{sin}\theta−\mathrm{cos}\theta\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{1}−\mathrm{sin}\left(\mathrm{2}\theta\right)…
Question Number 172608 by Mikenice last updated on 29/Jun/22 Answered by FelipeLz last updated on 03/Jul/22 $${f}\left({x}\right)\:=\:{x}−\mathrm{sin}\left({x}\right) \\ $$$$\int\frac{{x}\mathrm{cos}\left({x}\right)−\mathrm{sin}\left({x}\right)}{\left[{x}−\mathrm{sin}\left({x}\right)\right]^{\mathrm{2}} }{dx} \\ $$$$\int\frac{{x}\mathrm{cos}\left({x}\right)−{x}+{x}−\mathrm{sin}\left({x}\right)}{\left[{x}−\mathrm{sin}\left({x}\right)\right]^{\mathrm{2}} }{dx} \\ $$$$\int\frac{\left[{x}−\mathrm{sin}\left({x}\right)\right]−{x}\left[\mathrm{1}−\mathrm{cos}\left({x}\right)\right]}{\left[{x}−\mathrm{sin}\left({x}\right)\right]^{\mathrm{2}}…
Question Number 107018 by bemath last updated on 08/Aug/20 $$\mathcal{G}{iven}\:{f}\left({x}\right)=\frac{\mathrm{10}}{\mathrm{2}−\mathrm{sin}\:\mathrm{2}{x}}.\:{Find}\:{maximum}\:{value}\: \\ $$$${f}\left({x}\right). \\ $$ Commented by PRITHWISH SEN 2 last updated on 08/Aug/20 $$\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{to}\:\mathrm{be}\:\mathrm{max}.\:\mathrm{2}−\mathrm{sin}\:\mathrm{2x}\:\mathrm{must}\:\mathrm{be}\:\mathrm{min}.\: \\…