Question Number 186125 by Mingma last updated on 01/Feb/23 Answered by som(math1967) last updated on 01/Feb/23 $${let}\:\mathrm{2}^{{sin}\alpha} =\mathrm{3}^{{sin}\beta} =\mathrm{5}^{{sin}\theta} ={k} \\ $$$$\Rightarrow{k}^{\frac{\mathrm{1}}{{sin}\alpha}} =\mathrm{2}\Rightarrow{k}^{{cosec}\alpha} =\mathrm{2} \\…
Question Number 186126 by Mingma last updated on 01/Feb/23 Commented by Frix last updated on 01/Feb/23 $$\mathrm{tan}\:\mathrm{2}\alpha\:=\frac{\mathrm{2tan}\:\alpha}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha}\:\Rightarrow \\ $$$$\mathrm{tan}\:\mathrm{4}\alpha\:=\mathrm{tan}\:\left(\mathrm{2}×\left(\mathrm{2}\alpha\right)\right)\:=\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\alpha\right)\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{6tan}^{\mathrm{2}} \:\alpha\:+\mathrm{tan}^{\mathrm{4}} \:\alpha} \\ $$…
Question Number 120551 by bobhans last updated on 01/Nov/20 $$\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{tan}\:\mathrm{x}}\:−\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{x}}\:=\:\mathrm{sin}\:\mathrm{2x} \\ $$ Answered by john santu last updated on 01/Nov/20 $${recall}\:{that}\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\frac{\mathrm{2tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}} \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{tan}\:{x}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:{x}}\:=\:\frac{\mathrm{2tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}}…
Question Number 186061 by Rupesh123 last updated on 31/Jan/23 Answered by HeferH last updated on 31/Jan/23 Commented by HeferH last updated on 31/Jan/23 $$\mathrm{sin}\:\mathrm{30}°\:=\:\frac{{PM}}{\mathrm{7}} \\…
Question Number 185960 by Michaelfaraday last updated on 30/Jan/23 $${solve}: \\ $$$$\left({sinx}\overset{{x}} {\right)}=\mathrm{3}{x} \\ $$$${find}\:{x} \\ $$ Answered by Frix last updated on 30/Jan/23 $$\mathrm{We}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate}.…
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Question Number 185861 by cortano1 last updated on 29/Jan/23 $$\:\:\:\:\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{6}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{13}}\right)=? \\ $$ Commented by Frix last updated on 29/Jan/23 $$\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{4}} \\ $$ Commented by cortano1…
Question Number 54755 by Knight last updated on 10/Feb/19 $${solve}\:{for}\:{x} \\ $$$${sin}\left[\mathrm{2}{cos}^{−\mathrm{1}} \left\{{cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0} \\ $$$$ \\ $$$$\left({x}\:=\:\mathrm{1}\mp\sqrt{\mathrm{2}\:}\:,\:\mp\mathrm{1}\:,−\mathrm{1}\mp\sqrt{\mathrm{2}}\:\:\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…
Question Number 185821 by Rupesh123 last updated on 28/Jan/23 Answered by qaz last updated on 28/Jan/23 $$\mathrm{5arctan}\:\frac{\mathrm{1}}{\mathrm{7}}+\mathrm{2arctan}\:\frac{\mathrm{3}}{\mathrm{79}}=\mathrm{5}{Arg}\left(\mathrm{7}+{i}\right)+\mathrm{2}{Arg}\left(\mathrm{79}+\mathrm{3}{i}\right) \\ $$$$={Arg}\left[\left(\mathrm{7}+{i}\right)^{\mathrm{5}} \centerdot\left(\mathrm{79}+\mathrm{3}{i}\right)^{\mathrm{2}} \right]={Arg}\left[\mathrm{78125000}\left(\mathrm{1}+{i}\right)\right] \\ $$$$=\mathrm{arctan}\:\frac{\mathrm{78125000}}{\mathrm{78125000}}=\frac{\pi}{\mathrm{4}} \\ $$…
Question Number 185794 by Rupesh123 last updated on 27/Jan/23 Commented by Rupesh123 last updated on 27/Jan/23 Solve in R Answered by a.lgnaoui last updated on 27/Jan/23 $$\frac{\mathrm{tan}\:{x}}{\mathrm{tan}\:{y}}=−\mathrm{1}\:\:\:\:\:{y}=−{x}+{k}\pi\:\:\:\:\left({k}\in\mathbb{Z}\right)…