Question Number 166685 by cortano1 last updated on 25/Feb/22 $$\:\:\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{1}°+\mathrm{sec}\:^{\mathrm{2}} \mathrm{2}°+\mathrm{sec}\:^{\mathrm{2}} \mathrm{3}°+…+\mathrm{sec}\:^{\mathrm{2}} \mathrm{89}°=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 166686 by cortano1 last updated on 25/Feb/22 Commented by greogoury55 last updated on 26/Feb/22 $$\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\mathrm{10}°}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{40}°}\:+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{80}°}−\mathrm{2} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{80}°}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{80}°}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{40}°}−\mathrm{2} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{80}°\right)}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{80}}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{40}°}−\mathrm{2} \\…
Question Number 166683 by cortano1 last updated on 25/Feb/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 101116 by Jamshidbek2311 last updated on 30/Jun/20 Answered by 1549442205 last updated on 07/Jul/20 $$\boldsymbol{\mathrm{We}}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{tan}}\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{11}}+\mathrm{4}\boldsymbol{\mathrm{sin}}\:\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{11}}=\sqrt{\mathrm{11}} \\ $$$$\Leftrightarrow\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{11}}+\mathrm{4sin}\frac{\mathrm{2}\pi}{\mathrm{11}}\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{11}}=\sqrt{\mathrm{11}}\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{11}} \\ $$$$\Leftrightarrow\mathrm{3sin}\frac{\pi}{\mathrm{11}}−\mathrm{4sin}^{\mathrm{3}} \frac{\pi}{\mathrm{11}}+\mathrm{4}\left(\mathrm{2sin}\frac{\pi}{\mathrm{11}}\mathrm{cos}\frac{\pi}{\mathrm{11}}−\:\sqrt{\mathrm{11}}\:\right)\left(\:\mathrm{4cos}^{\mathrm{3}} \frac{\pi}{\mathrm{11}}−\mathrm{3cos}\frac{\pi}{\mathrm{11}}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{sin}\frac{\pi}{\mathrm{11}}\left[\mathrm{3}−\mathrm{4}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}}…
Question Number 101107 by PengagumRahasiamu last updated on 30/Jun/20 Commented by Dwaipayan Shikari last updated on 30/Jun/20 $${sin}\mathrm{6}°−{sin}\mathrm{66}°+{sin}\mathrm{78}°−{sin}\mathrm{42}° \\ $$$$=−\mathrm{2}{cos}\mathrm{36}°{sin}\mathrm{30}°+\mathrm{2}{cos}\mathrm{60}°{sin}\mathrm{18}° \\ $$$$={sin}\mathrm{18}°−{cos}\mathrm{36}°=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}−\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\…
Question Number 101040 by Rio Michael last updated on 30/Jun/20 $$\:\mathrm{Express}\:\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\mathrm{6}\theta\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\:\mathrm{sin}\:{A}\:−\:\mathrm{sin}\:{B} \\ $$$$\left({i}\right)\:\mathrm{using}\:\mathrm{that}\:\mathrm{result}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{2sin}\:\theta\left(\:\mathrm{cos}\:\mathrm{6}\theta\:+\:\mathrm{cos}\:\mathrm{4}\theta\:+\:\mathrm{cos}\:\mathrm{2}\theta\right)\:=\:\mathrm{sin}\:\mathrm{7}\theta−\mathrm{sin}\:\theta \\ $$$$\left({ii}\right)\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{result}\:\mathrm{cos}\:\left(\frac{\mathrm{12}\pi}{\mathrm{7}}\right)\:+\:\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)\:+\:\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({iii}\right)\:\mathrm{hence}\:\mathrm{find}\:\mathrm{a}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{to}\:\frac{\mathrm{sin7}\theta\:−\:\mathrm{sin}\:\theta}{\mathrm{cos}\:\mathrm{6}\theta\:+\:\mathrm{cos}\:\mathrm{4}\theta\:+\:\mathrm{cos}\:\mathrm{2}\theta}\:=\:\mathrm{1} \\ $$ Answered by maths mind last updated…
Question Number 100988 by bobhans last updated on 29/Jun/20 $$\mathrm{4sin}\:^{\mathrm{2}} {x}\:+\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{3}\: \\ $$$${find}\:{solution}\:{set}\:{on}\:{x}\in\left(\mathrm{0},\mathrm{2}\pi\right) \\ $$ Commented by Dwaipayan Shikari last updated on 29/Jun/20 $${sin}^{\mathrm{2}} {x}+\mathrm{2}{sinxcosx}+{cos}^{\mathrm{2}}…
Question Number 35362 by 26488679 last updated on 18/May/18 Commented by 26488679 last updated on 18/May/18 $${is}\:{arcsin}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:{can}\:{be}\:\:{equal}\:{to}\:\mathrm{210}°? \\ $$ Commented by ajfour last updated on…
Question Number 100870 by bramlex last updated on 29/Jun/20 Commented by bemath last updated on 29/Jun/20 $$\pi−\mathrm{e}\:=\:\mathrm{0}.\mathrm{4233} \\ $$$$\left(\pi−\mathrm{e}\right)^{\mathrm{ln}\left(\mathrm{1}−\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}\right)} \:\geqslant\:\left(\pi−\mathrm{e}\right)^{\mathrm{0}} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\Rightarrow\mathrm{ln}\left(\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x}\right)\right)\leqslant\:\mathrm{0} \\ $$$$\mathrm{ln}\left(−\mathrm{cos}\:\mathrm{2x}\right)\:\leqslant\:\mathrm{0}\:\Rightarrow−\mathrm{cos}\:\mathrm{2x}\:\leqslant\:\mathrm{1}…
Question Number 166360 by mathlove last updated on 19/Feb/22 $$\frac{\mathrm{sin}\:\mathrm{10}{x}}{{sin}\:\mathrm{2}{x}}−\frac{\mathrm{cos}\:\mathrm{10}{x}}{\mathrm{cos}\:\mathrm{2}{x}}=? \\ $$ Answered by som(math1967) last updated on 19/Feb/22 $$\frac{\boldsymbol{{sin}}\mathrm{10}\boldsymbol{{xcos}}\mathrm{2}\boldsymbol{{x}}−\boldsymbol{{cos}}\mathrm{10}\boldsymbol{{xsin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{xcos}}\mathrm{2}\boldsymbol{{x}}} \\ $$$$=\frac{\boldsymbol{{sin}}\left(\mathrm{10}\boldsymbol{{x}}−\mathrm{2}\boldsymbol{{x}}\right)}{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{xcos}}\mathrm{2}\boldsymbol{{x}}}=\frac{\mathrm{2}\boldsymbol{{sin}}\mathrm{8}\boldsymbol{{x}}}{\mathrm{2}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{xcos}}\mathrm{2}\boldsymbol{{x}}} \\ $$$$=\frac{\mathrm{4}\boldsymbol{{sin}}\mathrm{4}\boldsymbol{{xcos}}\mathrm{4}\boldsymbol{{x}}}{\boldsymbol{{sin}}\mathrm{4}\boldsymbol{{x}}}=\mathrm{4}\boldsymbol{{cos}}\mathrm{4}\boldsymbol{{x}} \\…