Question Number 170503 by libaolin last updated on 25/May/22 $$\mathrm{tan90}°=? \\ $$ Commented by mr W last updated on 25/May/22 $${you}\:{repeat}\:{the}\:{same}\:{question}\:{till} \\ $$$${you}\:{have}\:{got}\:{a}\:{wrong}\:{answer}? \\ $$$${when}\:{you}\:{look}\:{at}\:{the}\:{definition}\:{of}…
Question Number 39388 by maxmathsup by imad last updated on 05/Jul/18 $${calculate}\:{A}\:={tan}\left(\frac{\pi}{\mathrm{5}}\right).{tan}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right).{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right).{tan}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jul/18 $${tan}\mathrm{36}.{tan}\mathrm{72}.{tan}\mathrm{108}.{tan}\mathrm{144} \\ $$$${tan}\mathrm{108}={tan}\left(\mathrm{180}−\mathrm{72}\right)=−{tan}\mathrm{72} \\…
Question Number 170449 by cortano1 last updated on 24/May/22 $$\:\:\:{Solve}\:\frac{\mathrm{sin}\:\mathrm{12}°}{\mathrm{sin}\:\mathrm{24}°\:\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{sin}\:\mathrm{72}°}{\mathrm{sin}\:\left(\mathrm{36}°+{x}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170448 by libaolin last updated on 24/May/22 $$\mathrm{tan90}°= \\ $$ Commented by mokys last updated on 24/May/22 $${undefind} \\ $$ Terms of Service…
Question Number 170445 by cortano1 last updated on 24/May/22 $$\:\:\:{Solve}\:\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\left({x}+\mathrm{5}°\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2cos}\:\mathrm{55}°} \\ $$ Answered by thfchristopher last updated on 24/May/22 $$\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\left({x}+\mathrm{5}°\right)}=\frac{\mathrm{1}}{\mathrm{2cos}\:\mathrm{55}°} \\ $$$$\Rightarrow\mathrm{2sin}\:{x}\mathrm{cos}\:\mathrm{55}°=\mathrm{sin}\:\left({x}+\mathrm{5}°\right) \\ $$$$\mathrm{cos}\:\mathrm{55}° \\…
Question Number 104904 by bemath last updated on 24/Jul/20 $$\mathrm{4cos}\:^{\mathrm{2}} {x}\:\mathrm{sin}\:{x}\:−\mathrm{2sin}\:^{\mathrm{2}} {x}\:=\:\mathrm{3sin}\:{x} \\ $$$${where}\:−\frac{\pi}{\mathrm{2}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$ Answered by bramlex last updated on 24/Jul/20 $$\mathrm{sin}\:{x}\:\left(\mathrm{4cos}\:^{\mathrm{2}} {x}−\mathrm{2sin}\:{x}−\mathrm{3}\right)\:=\:\mathrm{0}…
Question Number 104894 by bramlex last updated on 24/Jul/20 $${prove}\:{that}\:\mathrm{cos}\:^{\mathrm{6}} {a}\:−\mathrm{sin}\:^{\mathrm{6}} {a}\:=\: \\ $$$$\mathrm{cos}\:\mathrm{2}{a}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{a}\right)\: \\ $$ Answered by john santu last updated on 24/Jul/20…
Question Number 170371 by cortano1 last updated on 22/May/22 $$\:{solve}\:\frac{\mathrm{sin}\:\left({x}+\mathrm{18}°\right)}{\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{sin}\:\mathrm{18}°}{\mathrm{sin}\:\mathrm{12}°} \\ $$ Answered by greougoury555 last updated on 22/May/22 $$\:\mathrm{sin}\:\mathrm{18}°=\mathrm{2sin}\:\mathrm{12}°\mathrm{sin}\:\mathrm{48}° \\ $$$$\:\frac{\mathrm{sin}\:\left({x}+\mathrm{18}°\right)}{\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{2sin}\:\mathrm{12}°\:\mathrm{sin}\:\mathrm{48}°}{\mathrm{sin}\:\mathrm{12}°} \\ $$$$\mathrm{sin}\:\left({x}+\mathrm{18}°\right)=\mathrm{2sin}\:{x}\:\mathrm{sin}\:\mathrm{48}° \\…
Question Number 104796 by mathocean1 last updated on 23/Jul/20 $${show}\:{that}\: \\ $$$$\mathrm{8}{cos}^{\mathrm{4}} {x}−\mathrm{8}{cos}^{\mathrm{2}} {x}+\mathrm{1}={cos}\mathrm{4}{x}. \\ $$ Commented by kaivan.ahmadi last updated on 23/Jul/20 $${cos}\mathrm{4}{x}=\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}{x}−\mathrm{1}=\mathrm{2}\left(\mathrm{2}{cos}^{\mathrm{2}}…
Question Number 104735 by bemath last updated on 23/Jul/20 Commented by 1549442205PVT last updated on 23/Jul/20 $$\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{question}}? \\ $$ Commented by bemath last updated on…