Question Number 14591 by Tinkutara last updated on 02/Jun/17 Answered by prakash jain last updated on 02/Jun/17 $$\mathrm{For}\:{x}\in\mathbb{R} \\ $$$$\mathrm{cosec}^{\mathrm{2}} {x}\geqslant\mathrm{1} \\ $$$$\mathrm{tan}^{\mathrm{100}} {x}\geqslant\mathrm{0} \\…
Question Number 14592 by Tinkutara last updated on 02/Jun/17 Commented by mrW1 last updated on 02/Jun/17 $$\mid\mathrm{cos}\:{x}\mid=\frac{{q}}{{p}}\leqslant\mathrm{1} \\ $$$$\mid\mathrm{cos}\:{y}\mid=\frac{{u}}{{v}}\leqslant\mathrm{1} \\ $$$$\mid\mathrm{cos}\:{z}\mid=\frac{\mathrm{1}}{\frac{{qu}}{{pv}}}=\frac{{pv}}{{qu}}\geqslant\mathrm{1}\leqslant\mathrm{1} \\ $$$${that}\:{means}\:{the}\:{absolute}\:{value}\:{of} \\ $$$$\mathrm{cos}\:{x},\:\mathrm{cos}\:{y}\:{and}\:\mathrm{cos}\:{z}\:{must}\:{be}\:\mathrm{1}.…
Question Number 14590 by Tinkutara last updated on 02/Jun/17 Answered by mrW1 last updated on 03/Jun/17 $$\mathrm{sin}\:{y}\geqslant\mathrm{sin}\:{x}−\mathrm{cos}\:\alpha\:\mathrm{cos}\:{x} \\ $$$$\mathrm{sin}\:{y}\geqslant\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}×\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}}×\mathrm{sin}\:{x}−\frac{\mathrm{cos}\:\alpha}{\:\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}}\:×\mathrm{cos}\:{x}\right) \\ $$$$\mathrm{sin}\:{y}\geqslant\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha}\left(\mathrm{cos}\:\theta×\mathrm{sin}\:{x}−\mathrm{sin}\:\theta\:×\mathrm{cos}\:{x}\right)…
Question Number 80113 by jagoll last updated on 31/Jan/20 $${how}\:{do}\:{you}\:{simply} \\ $$$$\mathrm{sin}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}{x}\right)+\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\right)\:? \\ $$ Commented by john santu last updated on 31/Jan/20 $${let}\:{x}=\:\mathrm{cos}\:{t}\:,\:\mathrm{3}{x}=\mathrm{tan}\:{y}…
Question Number 14470 by Tinkutara last updated on 01/Jun/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solution}\left(\mathrm{s}\right)\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{sin}\:{x}\:=\:\mathrm{0},\:{x}\:\in\:\left[\mathrm{0},\:\pi\right] \\ $$ Commented by mrW1 last updated on 01/Jun/17 $${since}\:{x}\:\in\:\left[\mathrm{0},\:\pi\right] \\ $$$${x}^{\mathrm{2}}…
Question Number 14467 by Tinkutara last updated on 01/Jun/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17 $$\mathrm{3}{sin}\mathrm{2}{A}=\mathrm{2}{sin}\mathrm{2}{B}\Rightarrow\mathrm{6}{sinAcosA}=\mathrm{4}{sinB}.{cosB} \\ $$$$\mathrm{9}{sin}^{\mathrm{2}} {Acos}^{\mathrm{2}} {A}=\mathrm{4}{sin}^{\mathrm{2}} {Bcos}^{\mathrm{2}} {B}\:\:\:\left({sinA}={t}\right) \\ $$$$\mathrm{9}{t}^{\mathrm{2}}…
Question Number 14438 by ajfour last updated on 31/May/17 $$\boldsymbol{{x}}=\frac{\mathrm{2}\boldsymbol{{a}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\theta},\:\boldsymbol{{y}}=\frac{\mathrm{2}\boldsymbol{{b}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\phi},\:{and} \\ $$$$\boldsymbol{{z}}=\frac{\mathrm{2}\boldsymbol{{c}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\psi}\:;\:{where}\:\boldsymbol{{a}},\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}} \\ $$$${are}\:{sides}\:{of}\:\bigtriangleup{ABC}\:{such}\:{that} \\ $$$$\boldsymbol{\phi}−\boldsymbol{\psi}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{A}}, \\ $$$$\boldsymbol{\psi}−\boldsymbol{\theta}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{B}},\:{and} \\ $$$$\boldsymbol{\theta}−\boldsymbol{\psi}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{C}}\:. \\ $$$${Find}\:{at}\:{least}\:{one}\:{feasible} \\ $$$${solution}\:{set}\:{of}\:\boldsymbol{\theta},\boldsymbol{\phi},\:{and}\:\boldsymbol{\psi}\:{in} \\…
Question Number 14366 by RasheedSindhi last updated on 31/May/17 $$\mathrm{Determine} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{cos}\left(\mathrm{sin}^{-\mathrm{1}} \mathrm{x}\right) \\ $$$$\left.\mathrm{b}\right)\:\mathrm{sin}\left(\mathrm{cos}^{-\mathrm{1}} \mathrm{x}\right) \\ $$$$\left.\mathrm{c}\right)\:\mathrm{cos}^{-\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\left.\mathrm{d}\right)\:\mathrm{sin}^{-\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right) \\ $$$$\left.\mathrm{e}\right)\:\mathrm{sin}^{-\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right)…
Question Number 14330 by myintkhaing last updated on 01/Jun/17 $${tan}\theta+{tan}\mathrm{2}\theta={tan}\mathrm{3}\theta \\ $$$$\frac{{tan}\theta+{tan}\mathrm{2}\theta}{\mathrm{1}−{tan}\theta{tan}\mathrm{2}\theta}\left(\mathrm{1}−{tan}\theta{tan}\mathrm{2}\theta\right)={tan}\mathrm{3}\theta \\ $$$${tan}\mathrm{3}\theta\left(\mathrm{1}−{tan}\theta{tan}\mathrm{2}\theta\right)={tan}\mathrm{3}\theta \\ $$$${tan}\theta{tan}\mathrm{2}\theta{tan}\mathrm{3}\theta\:\mathrm{0} \\ $$$$\mathrm{3}\theta={k}\pi,\:{where}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},… \\ $$$$\theta=\frac{{k}\pi}{\mathrm{3}} \\ $$$${tan}\theta=\mathrm{0}\:\:{or}\:\:{tan}\mathrm{2}\theta=\mathrm{0}\:{or}\:{tan}\:\mathrm{3}\theta\:=\:\mathrm{0} \\ $$$$\theta={m}\pi,\:{where}\:{m}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},… \\…
Question Number 79856 by Pratah last updated on 28/Jan/20 Commented by Pratah last updated on 28/Jan/20 $$\mathrm{proof}\:\mathrm{of}\:\mathrm{identity} \\ $$ Commented by mr W last updated…