Question Number 111513 by john santu last updated on 04/Sep/20 $$\:\:\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:^{\mathrm{2}} {x}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{sin}\:^{\mathrm{2}} {x}}\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{2}} \\ $$$${find}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:? \\ $$ Answered by bemath last updated on 04/Sep/20…
Question Number 111426 by weltr last updated on 03/Sep/20 Answered by Dwaipayan Shikari last updated on 03/Sep/20 $$\mathrm{4}\left({sin}^{\mathrm{6}} \alpha+{cos}^{\mathrm{6}} \alpha\right)−\mathrm{3}{cos}^{\mathrm{2}} \mathrm{2}\alpha \\ $$$$=\mathrm{4}\left({sin}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \alpha\right)^{\mathrm{3}}…
Question Number 111427 by weltr last updated on 03/Sep/20 $$\mathrm{4sin}\:^{\mathrm{6}} \alpha\:+\:\mathrm{4cos}\:^{\mathrm{6}} \alpha\:−\:\mathrm{3cos}\:^{\mathrm{2}} \mathrm{2}\alpha \\ $$ Commented by Dwaipayan Shikari last updated on 03/Sep/20 Question no 111426 Terms…
Question Number 111391 by Aina Samuel Temidayo last updated on 03/Sep/20 $$\mathrm{Compute}\:\mathrm{cos}\frac{\Pi}{\mathrm{12}} \\ $$ Answered by bemath last updated on 03/Sep/20 $$\frac{\pi}{\mathrm{6}}\:=\:\mathrm{2}×\frac{\pi}{\mathrm{12}}\:\Rightarrow\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\:=\:\mathrm{cos}\:\left(\mathrm{2}.\frac{\pi}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:=\:\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)−\mathrm{1}…
Question Number 111382 by john santu last updated on 03/Sep/20 Commented by kaivan.ahmadi last updated on 03/Sep/20 $${sec}^{\mathrm{2}} {x}=\mathrm{12}+{secx}\Rightarrow{sec}^{\mathrm{2}} {x}−{secx}−\mathrm{12}=\mathrm{0}\Rightarrow \\ $$$$\left({secx}−\mathrm{4}\right)\left({secx}+\mathrm{3}\right)=\mathrm{0}\Rightarrow{secx}=\mathrm{4},\:{secx}=−\mathrm{3} \\ $$$${by}\:{hypothesis}\:{secx}>\mathrm{0}\:{so}\:{we}\:{have}\:{secx}=\mathrm{4} \\…
Question Number 45842 by gunawan last updated on 17/Oct/18 Commented by maxmathsup by imad last updated on 17/Oct/18 $${what}\:{s}\:{this}\:{language}\:{sir}\:{joel}? \\ $$ Commented by MJS last…
Question Number 45794 by Tawa1 last updated on 16/Oct/18 Answered by MJS last updated on 16/Oct/18 $${a}\ast{c}={b}\:\Rightarrow\:{b}\ast{c}^{−\mathrm{1}} ={a}\:\Rightarrow\:{c}^{−\mathrm{1}} ={b} \\ $$$${b}\ast{c}={d}\:\Rightarrow\:{d}\ast{c}^{−\mathrm{1}} ={b}\:\Rightarrow\:{c}^{−\mathrm{1}} ={b} \\ $$$${c}\ast{c}={a}\:\Rightarrow\:{a}\ast{c}^{−\mathrm{1}}…
Question Number 45783 by habtetacke last updated on 16/Oct/18 $${sec}\frac{\pi}{\mathrm{7}}=\frac{\mathrm{48}−\sqrt{\mathrm{3}{a}}}{\mathrm{36}} \\ $$ Commented by MJS last updated on 17/Oct/18 $$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{question}? \\ $$ Terms of Service…
Question Number 45777 by habtetacke last updated on 16/Oct/18 $$\mathrm{2cos}\:\frac{\mathrm{3}\pi}{\mathrm{7}}=\frac{\mathrm{24}+\sqrt{\mathrm{3}{a}}}{\mathrm{72}}{where}\:{a}=−\mathrm{27}+\mathrm{4}\sqrt{\mathrm{147}.\mathrm{4}} \\ $$ Commented by MJS last updated on 17/Oct/18 $$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{question}? \\ $$ Terms of Service…
Question Number 45774 by habtetacke last updated on 16/Oct/18 $${trigonometruc}\:{valyes} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com