Question Number 30235 by NECx last updated on 18/Feb/18 $${find}\:{the}\:{sum}\:{of}\:{the}\:{infinite} \\ $$$${series}\: \\ $$$$\:\:\:\:\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right) \\ $$ Commented by prof Abdo imad last updated…
Question Number 161254 by cortano last updated on 15/Dec/21 $$\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}=? \\ $$ Commented by bobhans last updated on 15/Dec/21 $$\mathcal{P}\:=\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{14}}\right)+\mathrm{1}}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{14}}\right)\left[\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)}{\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{14}}\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{14}}\right)}\right]} \\ $$$$\:\mathcal{P}\:=\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)} \\ $$$$\:\mathcal{P}\:=\:\frac{−\mathrm{2}\left(\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)\right)+\mathrm{2cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)} \\…
Question Number 95703 by i jagooll last updated on 27/May/20 $$\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}^{\mathrm{o}} }\:+\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{20}^{\mathrm{o}} }\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{40}^{\mathrm{o}} }\:=? \\ $$ Commented by john santu last updated…
Question Number 161181 by cortano last updated on 13/Dec/21 $$\:\:{x}=\mathrm{cot}^{−\mathrm{1}} \left(\sqrt{\mathrm{cos}\:\theta}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{cos}\:\theta}\right) \\ $$$$\:\mathrm{sin}\:{x}=? \\ $$ Commented by MJS_new last updated on 13/Dec/21 $$\mathrm{sin}\:\left(\mathrm{arctan}\:{a}\:−\mathrm{arctan}\:{b}\right)\:=\frac{{a}−{b}}{\:\sqrt{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}}…
Question Number 30087 by rahul 19 last updated on 16/Feb/18 $$\mathrm{solve}: \\ $$$$\mathrm{cos3}{x}.{cos}^{\mathrm{3}} {x}+\mathrm{sin}\:\mathrm{3}{x}.\mathrm{sin}\:^{\mathrm{3}} {x}=\mathrm{0}. \\ $$ Answered by MJS last updated on 16/Feb/18 $${x}=\frac{\pi}{\mathrm{4}}+\frac{\pi\centerdot{z}}{\mathrm{2}};\:{z}\in\mathbb{Z}…
Question Number 30054 by rahul 19 last updated on 15/Feb/18 Answered by ajfour last updated on 16/Feb/18 $${let}\:\:\mathrm{sin}\:{x}=\:{s} \\ $$$$\mathrm{5}{s}^{\mathrm{2}} +\mathrm{4}{s}^{\mathrm{2}} \left(\mathrm{1}−{s}^{\mathrm{2}} \right)−\mathrm{4}\left(\mathrm{1}−\mathrm{2}{s}^{\mathrm{2}} \right)\:>\:\mathrm{0} \\…
Question Number 30000 by ajfour last updated on 14/Feb/18 $${If}\:\mathrm{cos}\:\alpha\:=\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\phi=\mathrm{sin}\:\gamma\:\mathrm{cos}\:\psi \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\beta\:=\:\mathrm{sin}\:\gamma\:\mathrm{sin}\:\psi\:=\mathrm{sin}\:\alpha\:\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\gamma\:=\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta\:=\mathrm{sin}\:\beta\:\mathrm{cos}\:\phi \\ $$$${then}\:{find}\:\:\mathrm{cos}\:\alpha,\:\mathrm{cos}\:\beta\:,\:\mathrm{cos}\:\gamma\:\:\: \\ $$$${briefly}\:{and}\:{if}\:{possible}\:{linearly} \\ $$$${in}\:{terms}\:{of}\:{only}\:\mathrm{sin}\:\theta,\:\mathrm{cos}\:\theta, \\ $$$$\mathrm{sin}\:\phi,\:\mathrm{cos}\:\phi,\:\mathrm{sin}\:\psi,\:\mathrm{cos}\:\psi\:. \\ $$ Commented…
Question Number 95531 by Abdulrahman last updated on 25/May/20 $$\frac{\mathrm{tanx}×\mathrm{ctg2x}}{\mathrm{tan}^{\mathrm{2}} \mathrm{x}−\mathrm{1}}=? \\ $$ Commented by PRITHWISH SEN 2 last updated on 25/May/20 $$\mathrm{put}\:\mathrm{cot2x}\:=\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{2tan}\:\mathrm{x}} \\…
Question Number 161060 by pete last updated on 11/Dec/21 $$\mathrm{Given}\:\mathrm{sin}\left(\mathrm{5x}−\mathrm{38}\right)=\mathrm{cos}\left(\mathrm{2x}+\mathrm{16}\right),\:\mathrm{0}°\leqslant\mathrm{x}\leqslant\mathrm{90}°, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$ Commented by cortano last updated on 11/Dec/21 $$\:{recall}\:\mathrm{sin}\:\left(\mathrm{90}°−{x}\right)=\:\mathrm{cos}\:{x} \\ $$$$\Leftrightarrow\:\mathrm{sin}\:\left(\mathrm{5}{x}−\mathrm{38}°\right)=\:\mathrm{sin}\:\left(\mathrm{90}°−\left(\mathrm{128}°−\mathrm{5}{x}\right)\right)=\mathrm{cos}\:\left(\mathrm{128}°−\mathrm{5}{x}\right) \\…
Question Number 95524 by Abdulrahman last updated on 25/May/20 $$\frac{\mathrm{1}}{\mathrm{sin10}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{cos10}}=? \\ $$ Commented by PRITHWISH SEN 2 last updated on 25/May/20 $$\frac{\mathrm{cos}\:\mathrm{10}−\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{10}}{\mathrm{sin}\:\mathrm{10cos}\:\mathrm{10}}\:=\:\mathrm{4}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{10}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{10}\right)}{\mathrm{2sin}\:\mathrm{10cos}\:\mathrm{10}} \\ $$$$=\mathrm{4}.\frac{\mathrm{sin}\:\mathrm{30}.\mathrm{cos}\:\mathrm{10}−\mathrm{cos}\:\mathrm{30}.\mathrm{sin}\:\mathrm{10}}{\mathrm{sin}\:\mathrm{20}}\:=\mathrm{4}.\frac{\mathrm{sin}\:\left(\mathrm{30}−\mathrm{10}\right)}{\mathrm{sin}\:\mathrm{20}} \\…