Category: Trigonometry

Question Number 35654 by chakraborty ankit last updated on 21/May/18 $${if}\:\:\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{tan}\:^{\mathrm{2}} \emptyset\:\:{Then}\:{proof}\:{that} \\ $$$$\mathrm{2cos}\:^{\mathrm{2}} \emptyset−\mathrm{1}=\mathrm{cos}\:^{\mathrm{2}} \emptyset−\mathrm{sin}\:^{\mathrm{2}} \emptyset=\mathrm{2tan}\:^{\mathrm{2}} \theta \\ $$ Answered by tanmay.chaudhury50@gmail.com…

Question Number 166718 by mathlove last updated on 26/Feb/22 $${csc}\mathrm{10}−\sqrt{\mathrm{3}}{sec}\mathrm{10}=? \\ $$ Commented by cortano1 last updated on 26/Feb/22 $$\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{10}°}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{cos}\:\mathrm{10}°}\:=\:\frac{\mathrm{cos}\:\mathrm{10}°−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{10}°}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:=\:\frac{\mathrm{2}\left(\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{10}°−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:=\frac{\mathrm{4}\left(\mathrm{cos}\:\mathrm{60}°\:\mathrm{cos}\:\mathrm{10}°−\mathrm{sin}\:\mathrm{60}°\:\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\…

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}}…