Question Number 109818 by Study last updated on 25/Aug/20 $${tanx}+{tan}\mathrm{2}{x}=\mathrm{1} \\ $$$${find}\:{the}\:{general}\:{solution} \\ $$ Commented by Dwaipayan Shikari last updated on 25/Aug/20 $${x}={k}\pi+\frac{\mathrm{261}}{\mathrm{720}}\pi\:\:\left({approximately}\right)\:\:\left({k}\in\mathbb{Z}\right) \\ $$…
Question Number 109715 by bobhans last updated on 25/Aug/20 $$\begin{cases}{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{sin}\:\mathrm{2}{y}=\frac{\mathrm{5}}{\mathrm{4}}}\\{\mathrm{cos}\:\left({x}−{y}\right)=\mathrm{2sin}\:\left({x}+{y}\right)}\end{cases}{where}\:\mathrm{0}<{x},{y}<\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \left({x}+{y}\right)\:=\:? \\ $$$$\:\:\:\:\:\:\bigtriangleup\frac{\flat{o}\flat}{{hans}}\bigtriangledown \\ $$ Answered by nimnim last updated on 25/Aug/20 $$\Rightarrow\mathrm{2sin}\left({x}+{y}\right)\mathrm{cos}\left({x}−{y}\right)=\frac{\mathrm{5}}{\mathrm{4}}…
Question Number 44172 by rahul 19 last updated on 22/Sep/18 $${The}\:{number}\:{of}\:{solutions}\:{of}\:{the}\:{equation} \\ $$$$\mathrm{cos}^{−\mathrm{1}} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{2}\pi}{\mathrm{3}}. \\ $$ Commented by tanmay.chaudhury50@gmail.com…
Question Number 44159 by rahul 19 last updated on 22/Sep/18 $${If}\:\mathrm{sin}^{−\mathrm{1}} {x}\:+\:\mathrm{sin}^{−\mathrm{1}} {y}\:+\:\mathrm{sin}^{−\mathrm{1}} \boldsymbol{{z}}\:=\pi\: \\ $$$${prove}\:{that}\:: \\ $$$$\left.{a}\right)\:{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\:{y}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\:+\boldsymbol{{z}}\sqrt{\mathrm{1}−\boldsymbol{{z}}^{\mathrm{2}} }=\:\mathrm{2}{xy}\boldsymbol{{z}} \\ $$$$\left.{b}\right)\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +\boldsymbol{{z}}^{\mathrm{4}}…
Question Number 44097 by peter frank last updated on 21/Sep/18 Answered by tanmay.chaudhury50@gmail.com last updated on 22/Sep/18 $${y}=\frac{\mathrm{2}\theta}{\mathrm{1}+\theta^{\mathrm{2}} } \\ $$$${let} \\ $$$${tanh}^{−\mathrm{1}} {y}={p} \\…
Question Number 44098 by peter frank last updated on 21/Sep/18 Answered by $@ty@m last updated on 21/Sep/18 $$\mathrm{cos}\:^{\mathrm{2}} {C}−\mathrm{cos}\:^{\mathrm{2}} {D} \\ $$$$=\left(\mathrm{cos}\:{C}+\mathrm{cos}\:{D}\right)\left(\mathrm{cos}\:{C}−\mathrm{cos}\:{D}\right) \\ $$$$=\mathrm{2cos}\:\frac{{C}+\mathrm{D}}{\mathrm{2}}\mathrm{cos}\:\frac{{C}−{D}}{\mathrm{2}}.\mathrm{2sin}\:\frac{{C}+{D}}{\mathrm{2}}\mathrm{sin}\:\frac{{D}−{C}}{\mathrm{2}} \\…
Question Number 109585 by nimnim last updated on 24/Aug/20 $${If}\:\left({xy}+{yz}+{zx}\right)=\mathrm{1},\:{then}\:{prove}\:{that} \\ $$$$\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{y}}{\mathrm{1}−{y}^{\mathrm{2}} }+\frac{{z}}{\mathrm{1}−{z}^{\mathrm{2}} }=\frac{\mathrm{4}{xyz}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)\left(\mathrm{1}−{z}^{\mathrm{2}} \right)} \\ $$ Answered by som(math1967) last updated…
Question Number 44028 by pieroo last updated on 20/Sep/18 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos2x}\right) \\ $$ Commented by maxmathsup by imad last updated on 20/Sep/18 $${by}\:{scalar}\:{product}\:{we}\:{can}\:{prove}\:{that}\:{cos}\left({a}+{b}\right)={cos}\:{a}\:{cosb}\:−{sina}\:{sinb} \\ $$$${let}\:{take}\:{a}={b}={x}\:\Rightarrow{cos}\left(\mathrm{2}{x}\right)={cos}^{\mathrm{2}}…
Question Number 109493 by bemath last updated on 24/Aug/20 Answered by Dwaipayan Shikari last updated on 24/Aug/20 $${cos}^{\mathrm{2}} \mathrm{2}{x}+{cos}^{\mathrm{2}} \mathrm{3}{x}+{cos}^{\mathrm{2}} \mathrm{4}{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{1}+{cos}\mathrm{4}{x}+\mathrm{1}+{cos}\mathrm{6}{x}+\mathrm{1}+{cos}\mathrm{8}{x}=\mathrm{3} \\ $$$${cos}\mathrm{4}{x}+{cos}\mathrm{6}{x}+{cos}\mathrm{8}{x}=\mathrm{0}…
Question Number 109461 by 150505R last updated on 23/Aug/20 Answered by 1549442205PVT last updated on 24/Aug/20 $$\mathrm{Applying}\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{AM}−\mathrm{GM}\:\mathrm{we} \\ $$$$\mathrm{have}:\left(\mathrm{2a}\right)^{\mathrm{2}} =\left(\mathrm{a}.\mathrm{tan}\alpha+\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}.\mathrm{tan}\beta+\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}.\mathrm{tan}\gamma\right)^{\mathrm{2}} \\ $$$$\leqslant\left[\left(\mathrm{a}^{\mathrm{2}} +\left(\sqrt{\mathrm{a}^{\mathrm{2}}…