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} \\ $$$$\mathrm{have}:\left(\mathrm{2a}\right)^{\mathrm{2}} =\left(\mathrm{a}.\mathrm{tan}\alpha+\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}.\mathrm{tan}\beta+\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}.\mathrm{tan}\gamma\right)^{\mathrm{2}} \\ $$$$\leqslant\left[\left(\mathrm{a}^{\mathrm{2}} +\left(\sqrt{\mathrm{a}^{\mathrm{2}}…
Question Number 43877 by peter frank last updated on 16/Sep/18 $$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}=\mathrm{cosec}\:−\mathrm{sin}\:{x}\:{prove}\:{that} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}=\left(-\mathrm{2}\pm\sqrt{\mathrm{5}}\right) \\ $$ Answered by ajfour last updated on 16/Sep/18 $$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\mathrm{sin}\:{x} \\…
Question Number 109401 by john santu last updated on 23/Aug/20 $$\:\:\:\:\:\:\:\frac{{JS}}{\_\_\mathrm{00\_00\_\_00}} \\ $$$${solve}\:{the}\:{equation}\:\mathrm{4sin}\:\mathrm{3}{x}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{3} \\ $$ Answered by mr W last updated on 23/Aug/20 $$\mathrm{12}\:\mathrm{sin}\:\mathrm{3}{x}+\mathrm{cos}\:\mathrm{3}{x}=\mathrm{9} \\…
Question Number 43847 by Rauny last updated on 16/Sep/18 $$\mathrm{I}\:\mathrm{think}\:\mathrm{tan}\:\mathrm{90}°=\mathrm{1}+{i}. \\ $$$$\because\mathrm{tan}\:{x}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\centerdot\frac{\mathrm{2}}{{e}^{{ix}} +{e}^{−{ix}} } \\ $$$${e}^{{ix}} :={E}, \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}=\frac{{E}−{E}^{−\mathrm{1}} }{\left({E}+{E}^{−\mathrm{1}} \right){i}}=−\frac{{E}−{E}^{−\mathrm{1}}…
Question Number 43810 by gyugfeet last updated on 15/Sep/18 $$\frac{\mathrm{1}−{cos}\theta+{co}\beta−{cos}\left(\theta+\beta\right)}{\mathrm{1}+{cos}\theta−{cos}\beta−{cos}\left(\theta+\beta\right)}={tan}\frac{\theta}{\mathrm{2}}.\:{cot}\:\frac{\beta}{\mathrm{2}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 15/Sep/18 $$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}+\beta\right){sin}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)−\mathrm{2}{cos}\left(\frac{\theta}{\mathrm{2}}+\beta\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$={tan}\frac{\theta}{\mathrm{2}}×\frac{{sin}\frac{\theta}{\mathrm{2}}+{sin}\left(\frac{\theta}{\mathrm{2}}+\beta\right)}{{cos}\left(\frac{\theta}{\mathrm{2}}\right)−{cos}\left(\frac{\theta}{\mathrm{2}}+\beta\right)} \\…
Question Number 43792 by gunawan last updated on 15/Sep/18 $$\mathrm{For}\:\pi\leqslant\theta<\mathrm{2}\pi \\ $$$$\mathrm{given}\: \\ $$$$\mathrm{P}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{3}\theta+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{4}\theta+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{cos}\:\mathrm{5}\theta \\ $$$$−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{sin}\:\mathrm{6}\theta−\frac{\mathrm{1}}{\mathrm{128}}\mathrm{cos}\:\mathrm{7}\theta+\ldots \\ $$$$\mathrm{Q}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\:\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}\:\:\mathrm{3}\theta+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{cos}\:\:\mathrm{4}\theta−\frac{\mathrm{1}}{\mathrm{32}}\mathrm{sin}\:\:\mathrm{5}\theta \\ $$$$−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{cos}\:\mathrm{6}\theta+\frac{\mathrm{1}}{\mathrm{128}}\mathrm{sin}\:\mathrm{7}\theta+… \\ $$$$\mathrm{so}\:\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}}.\:\mathrm{After}\:\mathrm{sin}\:\theta=−\frac{\mathrm{m}}{\mathrm{n}},\:\mathrm{where}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n} \\ $$$$\mathrm{prime}\:\mathrm{relatif}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{m}+\mathrm{n}\:\mathrm{is}\ldots \\…