Question Number 192540 by peter frank last updated on 20/May/23 Commented by mr W last updated on 17/Jun/23 $${why}\:{did}\:{you}\:{repeat}\:{the}\:{same}\:{question} \\ $$$${Q}#\mathrm{178890}? \\ $$ Terms of…
Question Number 61470 by Rasheed.Sindhi last updated on 03/Jun/19 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{other}\:\mathrm{solution}\:\mathrm{besides} \\ $$$$\left\{\mathrm{x}=\mathrm{a},\mathrm{y}=\mathrm{b}\right\}\:\mathrm{or}\:\left\{\mathrm{x}=\mathrm{b},\mathrm{y}=\mathrm{a}\right\}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\:\:\:\:\mathrm{x}+\mathrm{y}=\mathrm{a}+\mathrm{b}\:\:\wedge\:\mathrm{x}^{\mathrm{7}} +\mathrm{y}^{\mathrm{7}} =\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} \:\:? \\ $$$$ \\ $$ Answered…
Question Number 192543 by peter frank last updated on 20/May/23 Answered by leodera last updated on 20/May/23 $$\Delta\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:\left({x}\right)}{\mathrm{sin}\:\left({x}\right)+\mathrm{cos}\:\left({x}\right)}{dx} \\ $$$$\Delta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:\left({x}\right)+\mathrm{cos}\:\left({x}\right)}{\mathrm{sin}\:\left({x}\right)+\mathrm{cos}\:\left({x}\right)}{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…
Question Number 127004 by Dastronomer last updated on 26/Dec/20 Commented by Tinku Tara last updated on 26/Dec/20 http://www.tinkutara.com/equationeditorhelp.html Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192542 by peter frank last updated on 20/May/23 Answered by Frix last updated on 20/May/23 $$\sqrt{\mathrm{i}}=\sqrt{\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} }=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i} \\ $$$$\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} \right)^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}} =\:\:\:\:\:\:\:\:\:\:\left[\left({a}^{{b}} \right)^{{c}}…
Question Number 192537 by cortano12 last updated on 20/May/23 $$\:\begin{cases}{\mathrm{tan}\:\left(\alpha+\mathrm{2}\beta\right)=\sqrt{\mathrm{1}+\mathrm{2k}}}\\{\mathrm{tan}\:^{\mathrm{2}} \left(\alpha+\beta\right)\left\{\mathrm{1}+\mathrm{k}\:\mathrm{tan}\:^{\mathrm{2}} \beta\right\}=\mathrm{k}+\mathrm{tan}\:^{\mathrm{2}} \beta}\end{cases} \\ $$$$\:\mathrm{Find}\:\mathrm{cot}\:\mathrm{2}\beta\:. \\ $$ Answered by a.lgnaoui last updated on 20/May/23 $$\:\:\:\:\mathrm{tan}\:\left[\left(\alpha+\beta\right)+\beta\right]=\sqrt{\mathrm{1}+\mathrm{2k}}\:\:\:\:\:\:\:\:\:\:\:…
Question Number 127002 by bramlexs22 last updated on 26/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}+\mathrm{cos}\:\left(\mathrm{3}{x}\right)−\mathrm{3csch}\:\left({x}\right)}{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=? \\ $$ Commented by Evimene last updated on 26/Dec/20 $$ \\ $$$$\:\: \\…
Question Number 192536 by cortano12 last updated on 20/May/23 $$\:\:\mathrm{R}=\frac{\mathrm{3sin}\:\mathrm{5}°+\mathrm{4cos}\:\mathrm{5}°−\mathrm{5cos}\:\mathrm{58}°+\mathrm{35}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{13}°}{\mathrm{cos}\:\mathrm{5}°}=? \\ $$ Answered by Tomal last updated on 20/May/23 $$\:\mathrm{R}=\frac{\mathrm{3sin}\:\mathrm{5}°+\mathrm{4cos}\:\mathrm{5}°−\mathrm{5cos}\:\mathrm{58}°+\mathrm{35}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{13}°}{\mathrm{cos}\:\mathrm{5}°}=? \\ $$$$\left.{R}=\frac{\left(\mathrm{3}×\mathrm{0}.\mathrm{0872}\right)+\left(\mathrm{4}×\mathrm{0}.\mathrm{996}\right)−\left(\mathrm{5}×\mathrm{0}.\mathrm{53}\right)+\left(\mathrm{35}\:\underbrace{\frown}\right.}{}\mathrm{2}×\mathrm{0}.\mathrm{974}\right) \\ $$ Terms…
Question Number 61465 by arcana last updated on 02/Jun/19 $$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({t}\right)+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({t}\right)}{dt}=\frac{\mathrm{2}\pi}{{ab}}? \\ $$ Commented by maxmathsup by imad last updated…
Question Number 61461 by aliesam last updated on 02/Jun/19 Answered by MJS last updated on 03/Jun/19 $$\mathrm{sin}\:{x}\:+\mathrm{sin}\:{y}\:={a} \\ $$$$\mathrm{cos}\:{x}\:+\mathrm{cos}\:{y}\:={b} \\ $$$${u}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\wedge\:{v}=\mathrm{tan}\:\frac{{y}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\left[\mathrm{sin}\:\mathrm{2arctan}\:{t}\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} };\:\mathrm{cos}\:\mathrm{2arctan}\:{t}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}}…