Question Number 166296 by mathlove last updated on 18/Feb/22 $$\left(\mathrm{cos}\:{x}\right)^{\mathrm{2022}} −\left(\mathrm{sin}\:{x}\right)^{\mathrm{2022}} =\mathrm{1} \\ $$$${x}=? \\ $$ Commented by mkam last updated on 18/Feb/22 $$\boldsymbol{{x}}\:=\:\boldsymbol{{n}\pi}\:\forall\:\boldsymbol{{n}}\:\in\:\boldsymbol{{Z}}\: \\…
Question Number 100695 by john santu last updated on 28/Jun/20 Commented by bobhans last updated on 28/Jun/20 $$\mathrm{sin}\:\mathrm{18}\:=\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\:,\:\mathrm{sin}\:\mathrm{54}\:=\:\mathrm{cos}\:\mathrm{36}\:=\:\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{18} \\ $$$$\mathrm{sin}\:\mathrm{54}\:=\:\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}\right)\:=\:\mathrm{1}−\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{4}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{18}}\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{54}}\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}\:−\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}+\mathrm{1}} \\ $$$$=\:\mathrm{4}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}−\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}\right)\:=\:\mathrm{2}…
Question Number 166223 by Fikret last updated on 15/Feb/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 166168 by alcohol last updated on 14/Feb/22 $${prove} \\ $$$$\underset{{r}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}\:+\:\left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right)\pi}\:=\:{tan}\left({x}\right) \\ $$$$\left(\underset{{r}=−\infty} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{x}\:+\:{r}}\right)\left(\underset{{r}=−\infty} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{x}\:+\:{r}}\right)\:=\:−\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\left(\:{r}\:=\:{odd}\right)\:\:\:\:\:\:\:\:\left({r}\:=\:{even}\right) \\ $$…
Question Number 166169 by cortano1 last updated on 14/Feb/22 $$\:\:\mathrm{sin}^{\mathrm{7}} \left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{3}} \left(\mathrm{x}\right)}=\mathrm{cos}\:^{\mathrm{7}} \left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{3}} \left(\mathrm{x}\right)} \\ $$$$ \\ $$ Answered by greogoury55 last updated on 14/Feb/22…
Question Number 166170 by mathls last updated on 14/Feb/22 $${tan}\left({a}+{b}\right)=\frac{\mathrm{1}}{\mathrm{17}}\:\:\:,\:\:\:{tan}\left({a}−{b}\right)=\frac{\mathrm{11}}{\mathrm{13}} \\ $$$${tan}\mathrm{2}{a}=?\:\:\:\:\:\:{tan}\mathrm{2}{b}=? \\ $$ Answered by cortano1 last updated on 14/Feb/22 $$\Rightarrow\mathrm{2}{a}\:=\:\left({a}+{b}\right)+\left({a}−{b}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{2}{a}\:=\frac{\frac{\mathrm{1}}{\mathrm{17}}+\frac{\mathrm{11}}{\mathrm{13}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{17}}.\frac{\mathrm{11}}{\mathrm{13}}}\:=\:\frac{\mathrm{20}}{\mathrm{21}} \\…
Question Number 100522 by bemath last updated on 27/Jun/20 Answered by bramlex last updated on 27/Jun/20 Commented by bramlex last updated on 27/Jun/20 $$\left(\mathrm{1}\right)\pi−\alpha\:+\:\pi−\beta\:+\:{x}\:+\:{y}\:=\:\mathrm{2}\pi \\…
Question Number 34980 by NECx last updated on 14/May/18 $${If}\:{cos}\mathrm{45}°\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:.\:{Find}\:{cos}\mathrm{45}.\mathrm{1}° \\ $$ Answered by ajfour last updated on 14/May/18 $$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{1800}}\right)\approx\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}\right)−\frac{\pi}{\mathrm{1800}}\mathrm{sin}\:\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}−\frac{\pi}{\mathrm{1800}}\right)\:=\mathrm{0}.\mathrm{707}\left(\mathrm{1}−\frac{\mathrm{0}.\mathrm{0314}}{\mathrm{18}}\right) \\ $$$$\:\:\:=\mathrm{0}.\mathrm{707}\left(\mathrm{1}−\mathrm{0}.\mathrm{00173}\right) \\…
Question Number 100491 by bobhans last updated on 27/Jun/20 Answered by bramlex last updated on 27/Jun/20 $${let}\:\mathrm{arccot}\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:{z}\:\Rightarrow\:\mathrm{cot}\:{z}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${let}\:\mathrm{arccot}\:\frac{\mathrm{2}}{\mathrm{3}}\:=\:{w}\:\Rightarrow\mathrm{cot}\:{w}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{tan}\:\left({z}+{w}\right)\:=\:\frac{\mathrm{tan}\:{z}+\mathrm{tan}\:{w}}{\mathrm{1}−\mathrm{tan}\:{z}.\mathrm{tan}\:{w}} \\ $$$$=\:\frac{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}}{\mathrm{1}−\mathrm{2}.\frac{\mathrm{3}}{\mathrm{2}}}\:=\:\frac{\mathrm{7}}{\mathrm{2}−\mathrm{6}}\:=\:−\frac{\mathrm{7}}{\mathrm{4}\:}\:\blacksquare\: \\ $$…
Question Number 166013 by mathlove last updated on 11/Feb/22 Answered by JDamian last updated on 11/Feb/22 $${x}_{{r}} ={e}^{{i}\frac{\pi}{\mathrm{2}^{{r}} }} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} \centerdot\centerdot\centerdot{x}_{\infty} =\:{e}^{{i}\pi\underset{{r}}…