Question Number 145305 by puissant last updated on 04/Jul/21 $$\mathrm{R}\acute {\mathrm{e}soudre}\:\:\:\frac{\frac{\mathrm{8}}{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}\:+\:\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}\:+\:\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right)}\:=\:\mathrm{cotan}^{\mathrm{2}} \left(\mathrm{x}\right)+\frac{\mathrm{4}}{\mathrm{3}} \\ $$ Commented by imjagoll last updated on 04/Jul/21 $$\mathrm{put}\:\mathrm{u}\:=\:\mathrm{cos}\:^{\mathrm{2}}…
Question Number 79766 by jagoll last updated on 28/Jan/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{x} \\ $$$$\mathrm{2}^{\mathrm{x}} =\left(\mathrm{1}+\mathrm{tan}\:\mathrm{0}.\mathrm{01}^{\mathrm{o}} \right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{0}.\mathrm{02}^{\mathrm{o}} \right) \\ $$$$\left(\mathrm{1}+\mathrm{tan}\:\mathrm{0}.\mathrm{03}^{\mathrm{o}} \right)…\left(\mathrm{1}+\mathrm{tan}\:\mathrm{44}.\mathrm{99}^{\mathrm{o}} \right) \\ $$ Answered by mr W…
Question Number 14189 by tawa tawa last updated on 29/May/17 $$\mathrm{Solve}:\:\:\:\sqrt{\mathrm{sinx}}\:+\:\sqrt{\mathrm{cosx}}\:=\:\mathrm{1} \\ $$$$\mathrm{x}\:\in\:\mathrm{R} \\ $$ Answered by ajfour last updated on 29/May/17 $${let}\:\sqrt{\mathrm{sin}\:{x}}={u}\:\:{and}\:\sqrt{\mathrm{cos}\:{x}}={v} \\ $$$${this}\:{means}\:\:{u}+{v}=\mathrm{1}\:\:\:\:\:\:\:…\left({i}\right)…
Question Number 14174 by Nayon last updated on 29/May/17 $$\:\:{xtan}\left({cos}^{−\mathrm{1}} {x}\right)={cos}\left({sin}^{−\mathrm{1}} {x}\right) \\ $$$${prove}\:{it}….. \\ $$ Answered by ajfour last updated on 29/May/17 $${let}\:\mathrm{cos}^{−\mathrm{1}} {x}=\theta\:\:\:\Rightarrow\:\:{x}=\mathrm{cos}\:\theta…
Question Number 145208 by imjagoll last updated on 03/Jul/21 $$\:\mathrm{If}\:\mathrm{P}\::\:\mathrm{Q}\:=\:\mathrm{tan}\:\mathrm{2A}\::\:\mathrm{cos}\:\mathrm{A} \\ $$$$\:\:\:\:\:\mathrm{Q}\::\:\mathrm{R}\:=\:\mathrm{cos}\:\mathrm{2A}\::\:\mathrm{sin}\:\mathrm{2A} \\ $$$$\:\:\:\:\:\mathrm{then}\:\mathrm{P}\::\:\mathrm{R}\:=? \\ $$ Answered by EDWIN88 last updated on 03/Jul/21 $$\:\frac{{P}}{{Q}}\:=\:\frac{\mathrm{2sin}\:{A}\:\mathrm{cos}\:{A}}{\mathrm{cos}\:\mathrm{2}{A}\:\mathrm{cos}\:{A}}\:=\:\frac{\mathrm{2sin}\:{A}}{\mathrm{cos}\:\mathrm{2}{A}} \\…
Question Number 14130 by Joel577 last updated on 28/May/17 $$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{if} \\ $$$$\frac{\mathrm{5}\:−\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{sin}\:\theta}\:\:\geqslant\:\mathrm{2}{k}\:\:\:\:\:\:\:\:\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\pi \\ $$ Commented by Joel577 last updated on 28/May/17 $$\frac{\mathrm{5}\:−\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{sin}\:\theta}\:\geqslant\:\mathrm{2}{k} \\ $$$$\frac{\mathrm{4}\:+\:\mathrm{2sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}\:\theta}\:\geqslant\:\mathrm{2}{k}…
Question Number 14078 by Tinkutara last updated on 27/May/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/May/17 $$\mathrm{4}{cos}^{\mathrm{3}} {x}−\mathrm{3}{cosx}−\sqrt{\mathrm{3}}{sinx}.{cosx}+{cos}^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{2}}=−\mathrm{2} \\ $$$$\mathrm{8}{cos}^{\mathrm{3}} {x}+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{2}\sqrt{\mathrm{3}}{sinx}.{cosx}−\mathrm{6}{cosx}+\mathrm{3}=\mathrm{0} \\ $$$${cosx}={t}\Rightarrow…
Question Number 14077 by Tinkutara last updated on 27/May/17 Answered by ajfour last updated on 28/May/17 $${let}\:\frac{\pi}{\mathrm{4}}\mathrm{cot}\:\theta=\alpha\:\:{and}\:\:\frac{\pi}{\mathrm{4}}\mathrm{tan}\:\theta=\beta \\ $$$$\:\:{then}\:\:\mathrm{sin}\:\alpha=\mathrm{cos}\:\beta={f}\:\:\left({say}\right) \\ $$$$\beta=\mathrm{2}{p}\pi\pm\mathrm{cos}^{−\mathrm{1}} {f}\:\:;\:\:{p}\:\in\:{Z}\:\:\:\:\:\:….\left({a}\right) \\ $$$$\alpha={m}\pi+\left(−\mathrm{1}\right)^{{m}} \mathrm{sin}^{−\mathrm{1}}…
Question Number 14030 by Tinkutara last updated on 27/May/17 $$\mathrm{The}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{equation} \\ $$$$\mathrm{tan}\:{x}\:\mathrm{tan}\:\mathrm{4}{x}\:=\:\mathrm{1}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\frac{\pi}{\mathrm{10}}\:,\:{n}\:\in\:{Z}\:−\:\left\{{n}\::\:{n}\:=\:\mathrm{5}{k}\:+\mathrm{2};\:{k}\:\in\:{Z}\right\} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{4}{n}\:−\:\mathrm{1}\right)\frac{\pi}{\mathrm{10}}\:,\:{n}\:\in\:{Z} \\ $$$$\left(\mathrm{3}\right)\:\frac{{n}\pi}{\mathrm{10}}\:,\:{n}\:\in\:{Z} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}{n}\pi\:+\:\frac{\pi}{\mathrm{10}}\:,\:{n}\:\in\:{Z} \\ $$ Answered by ajfour…
Question Number 144998 by imjagoll last updated on 01/Jul/21 $$\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{1}°+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}°+\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}°+…+\mathrm{cos}\:^{\mathrm{2}} \mathrm{360}°\:=\:? \\ $$ Answered by Dwaipayan Shikari last updated on 01/Jul/21 $$\frac{\mathrm{360}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{2}°+{cos}\mathrm{4}°+{cos}\mathrm{6}°+…+{cos}\mathrm{720}°\right)…