Question Number 207641 by hardmath last updated on 21/May/24 $$\mathrm{Find}:\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}}\:\:=\:\:? \\ $$ Commented by Frix last updated on 22/May/24 $$\mathrm{4}−\mathrm{2}\sqrt{\mathrm{7}}\mathrm{cos}\:\frac{\pi+\mathrm{2sin}^{−\mathrm{1}} \:\frac{\mathrm{37}\sqrt{\mathrm{7}}}{\mathrm{98}}}{\mathrm{6}} \\ $$ Answered…
Question Number 207639 by hardmath last updated on 21/May/24 Commented by Berbere last updated on 21/May/24 $${false}\:{not}\:{true}\: \\ $$$$\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\sim\frac{\mathrm{4}^{{n}} }{\:\sqrt{\pi{n}}} \\ $$$$\frac{\mathrm{2}^{−\mathrm{2}\left({n}−\mathrm{1}\right)} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{4}\right)}\begin{pmatrix}{{n}}\\{\mathrm{2}{n}}\end{pmatrix}^{\mathrm{2}} \sim\frac{\mathrm{4}^{{n}} }{\pi{n}}\:{diverge}…
Question Number 207638 by hardmath last updated on 21/May/24 Commented by TonyCWX08 last updated on 22/May/24 $${arcsin}\left({sin}\mathrm{5}\right)=\mathrm{5} \\ $$$${arccis}\left({cos}\mathrm{6}\right)=\mathrm{6} \\ $$$${arctan}\left({tan}\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{5}+\mathrm{6}+\mathrm{2}=\mathrm{13} \\ $$…
Question Number 207615 by mr W last updated on 21/May/24 $$\begin{cases}{{u}_{{n}+\mathrm{1}} =\frac{{au}_{{n}} +{b}}{{cu}_{{n}} +{d}}}\\{{u}_{\mathrm{0}} ={k}}\end{cases} \\ $$$${find}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$ Commented by mr W last…
Question Number 207597 by hardmath last updated on 19/May/24 Answered by A5T last updated on 20/May/24 Commented by A5T last updated on 20/May/24 $$\frac{{a}}{{x}}=\frac{{b}}{{y}}\Rightarrow\frac{{a}}{{b}}=\frac{{x}}{{y}}\Rightarrow\frac{{x}}{{x}+{y}}=\frac{{a}}{{a}+{b}} \\…
Question Number 207588 by hardmath last updated on 19/May/24 $$\sqrt{\mathrm{2}}\:\mathrm{sin}\boldsymbol{\mathrm{x}}\:\:+\:\:\mathrm{cos}\boldsymbol{\mathrm{x}}\:\:\geqslant\:\:\mathrm{1} \\ $$ Answered by mr W last updated on 19/May/24 $$\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{cos}\:{x}\geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{sin}\:\alpha\:\:\mathrm{sin}\:{x}+\mathrm{cos}\:\alpha\:\mathrm{cos}\:{x}\geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${with}\:\alpha=\mathrm{cos}^{−\mathrm{1}}…
Question Number 207590 by hardmath last updated on 19/May/24 $$\mathrm{x}^{\:\boldsymbol{\mathrm{lg}}\:\boldsymbol{\mathrm{x}}} \:\:<\:\:\mathrm{10}^{\mathrm{4}} \\ $$$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{roots}? \\ $$ Commented by hardmath last updated on 19/May/24 $$\mathrm{professor},\:\mathrm{integer}\:\mathrm{roots} \\ $$…
Question Number 207562 by hardmath last updated on 18/May/24 $$\mathrm{4}\:\mathrm{cos}\:\mathrm{50}°\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°}\:\:=\:\:? \\ $$ Commented by Frix last updated on 18/May/24 $$−\mathrm{2tan}\:\mathrm{10}° \\ $$ Terms of Service…
Question Number 207561 by hardmath last updated on 18/May/24 $$\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\boldsymbol{\mathrm{x}}\:\:+\:\:\mathrm{sin}\:\mathrm{2}\boldsymbol{\mathrm{x}}\:\:=\:\:\mathrm{2} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by mathzup last updated on 18/May/24 $${e}\Leftrightarrow\mathrm{4}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\:+{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}\:\Leftrightarrow \\ $$$$\mathrm{2}−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)+{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}\:\Leftrightarrow…
Question Number 207558 by hardmath last updated on 18/May/24 $$\left(\mathrm{x}\:−\:\mathrm{2}\right)\:\mathrm{lg}\:\frac{\mathrm{x}}{\mathrm{3}}\:\:\geqslant\:\:\mathrm{0} \\ $$ Answered by mr W last updated on 19/May/24 $${x}−\mathrm{2}\geqslant\mathrm{0}\:\wedge\:\frac{{x}}{\mathrm{3}}\geqslant\mathrm{1}\:\Rightarrow{x}\geqslant\mathrm{3}\:\checkmark \\ $$$${or} \\ $$$${x}−\mathrm{2}\leqslant\mathrm{0}\:\wedge\:\mathrm{0}<\frac{{x}}{\mathrm{3}}\leqslant\mathrm{1}\:\Rightarrow\mathrm{0}<{x}\leqslant\mathrm{2}\:\checkmark…