Question Number 161559 by stelor last updated on 19/Dec/21 $${please}\:{show}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:+\:{cosx}\:+\:{cos}\mathrm{2}{x}\:+\:{cos}\mathrm{3}{x}\:+\:…\:+\:{cosnx}\:=\:\frac{{sin}\left[\left({n}+\mathrm{1}\right)\frac{{x}}{\mathrm{2}}\right]}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}} \\ $$ Answered by Ar Brandon last updated on 19/Dec/21 $${P}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}{x}+\mathrm{cos2}{x}+\mathrm{cos3}{x}+\centerdot\centerdot\centerdot+\mathrm{cos}{nx} \\ $$$$\left(\mathrm{2sin}\frac{{x}}{\mathrm{2}}\right){P}=\mathrm{sin}\frac{{x}}{\mathrm{2}}+\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}{x}+\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos2}{x}+\centerdot\centerdot\centerdot+\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}{nx}…
Question Number 161521 by vvvv last updated on 19/Dec/21 $$\boldsymbol{\mathrm{x}}^{\mathrm{8}} +\boldsymbol{\mathrm{ax}}^{\mathrm{4}} +\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{{a}}=?\: \\ $$$$\boldsymbol{{is}}\:\:\boldsymbol{{x}}_{\mathrm{1}} +\boldsymbol{{x}}_{\mathrm{2}} +\boldsymbol{{x}}_{\mathrm{3}} +\boldsymbol{{x}}_{\mathrm{4}} =? \\ $$ Commented by mr…
Question Number 161516 by naka3546 last updated on 19/Dec/21 $${f}'\left({a}\right)\:\:{is}\:\:{derivative}\:\:{of}\:\:{function}\:\:{f}\left({a}\right)\:. \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{f}\left({a}−\mathrm{2}{h}^{\mathrm{2}} \right)−{f}\left({a}+{h}^{\mathrm{3}} \right)}{{h}^{\mathrm{2}} }\:\:=\:\:? \\ $$ Commented by cortano last updated on 19/Dec/21…
Question Number 161513 by Gbenga last updated on 19/Dec/21 $$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}} \\ $$$$−\mathrm{1}−\mathrm{1}/\mathrm{3}^{\mathrm{2}} −\mathrm{1}/\mathrm{5}^{\mathrm{2}} −\mathrm{1}/\mathrm{7}^{\mathrm{2}} −… \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161505 by naka3546 last updated on 18/Dec/21 $$\sqrt[{\mathrm{3}}]{{a}\:+\:\frac{{a}+\mathrm{8}}{\mathrm{3}}\:\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\:+\:\sqrt[{\mathrm{3}}]{{a}\:−\:\frac{{a}+\mathrm{8}}{\mathrm{3}}\:\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\:\:=\:\:? \\ $$ Commented by cortano last updated on 18/Dec/21 $$\:{x}=\sqrt[{\mathrm{3}}]{{a}+\frac{\left({a}+\mathrm{8}\right)}{\mathrm{3}}\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\:+\sqrt[{\mathrm{3}}]{{a}−\frac{\left({a}+\mathrm{8}\right)}{\mathrm{3}}\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}} \\ $$$$\:\Rightarrow\left({a}+\cancel{\left(\frac{{a}+\mathrm{8}}{\mathrm{3}}\right)\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\right)+\left({a}−\cancel{\left(\frac{{a}+\mathrm{8}}{\mathrm{3}}\right)\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\right)−{x}^{\mathrm{3}} =−\mathrm{3}{x}\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} −\left(\frac{{a}+\mathrm{8}}{\mathrm{3}}\right)^{\mathrm{2}} \left(\frac{{a}−\mathrm{1}}{\mathrm{3}}\right)}…
Question Number 161504 by mathocean1 last updated on 18/Dec/21 $${Montrer}\:\grave {{a}}\:{partir}\:{du}\:{crit}\grave {{e}re}\:{de}\: \\ $$$${Cauchy}\:{que}\:{U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:{est}\:{une} \\ $$$${de}\:{Cauchy}. \\ $$$$−−−−−−−−−−−−−−−− \\ $$$${Show}\:{by}\:{using}\:{Cauchy}'{s}\:{sequence} \\…
Question Number 161491 by wongb1506 last updated on 18/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161476 by MathsFan last updated on 18/Dec/21 $$\:{Between}\:\frac{\mathrm{3}}{\mathrm{6}}\:\:{and}\:−\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\:{How}\:{do}\:{i}\:{list}\:{two}\:{rational}\: \\ $$$$\:{numbers}\:{please}? \\ $$ Commented by mr W last updated on 18/Dec/21 $$\frac{\mathrm{3}}{\mathrm{6}}=\frac{\mathrm{15}}{\mathrm{30}}…
Question Number 161464 by ZiYangLee last updated on 18/Dec/21 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{in}\:\Delta\mathrm{ABC}, \\ $$$$\left(\mathrm{sin}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}\right):\left(\mathrm{sin}\:\mathrm{B}+\mathrm{sin}\:\mathrm{C}\right):\left(\mathrm{sin}\:\mathrm{C}+\mathrm{sin}\:\mathrm{A}\right)=\:\mathrm{6}:\:\mathrm{4}:\:\mathrm{5} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{A}. \\ $$ Commented by cortano last updated on 18/Dec/21 $$\:\begin{cases}{\mathrm{sin}\:{A}+\mathrm{sin}\:{B}=\mathrm{6}{k}}\\{\mathrm{sin}\:{B}+\mathrm{sin}\:{C}=\mathrm{4}{k}}\\{\mathrm{sin}\:{C}+\mathrm{sin}\:{A}=\mathrm{5}{k}}\end{cases} \\…
Question Number 30390 by mondodotto@gmail.com last updated on 21/Feb/18 Commented by abdo imad last updated on 21/Feb/18 $${we}\:{have}\:{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:+\mathrm{3}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}\:\:{tbe}\:{ch}\:.{x}+\mathrm{2}={cht}\:{give} \\ $$$${I}=\:\int\:\sqrt{{ch}^{\mathrm{2}} {t}\:−\mathrm{1}}\:{shtdt}\:=\:\int\:{sh}^{\mathrm{2}} {t}\:{dt}\:=\:\int\frac{\mathrm{1}−{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\…