Category: Trigonometry

Question Number 166695 by amin96 last updated on 25/Feb/22 Answered by som(math1967) last updated on 25/Feb/22 $$\left({x}^{{x}+\frac{\mathrm{1}}{{x}}} +\mathrm{1}+\mathrm{1}+{x}^{{x}+\frac{\mathrm{1}}{{x}}} \right) \\ $$$$={x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:+\mathrm{2}\:\:\:\left[{x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\right] \\ $$$$=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}}…

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 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 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 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…