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 \\ $$$$\mathrm{given}\: \\ $$$$\mathrm{P}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{3}\theta+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{4}\theta+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{cos}\:\mathrm{5}\theta \\ $$$$−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{sin}\:\mathrm{6}\theta−\frac{\mathrm{1}}{\mathrm{128}}\mathrm{cos}\:\mathrm{7}\theta+\ldots \\ $$$$\mathrm{Q}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\:\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}\:\:\mathrm{3}\theta+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{cos}\:\:\mathrm{4}\theta−\frac{\mathrm{1}}{\mathrm{32}}\mathrm{sin}\:\:\mathrm{5}\theta \\ $$$$−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{cos}\:\mathrm{6}\theta+\frac{\mathrm{1}}{\mathrm{128}}\mathrm{sin}\:\mathrm{7}\theta+… \\ $$$$\mathrm{so}\:\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}}.\:\mathrm{After}\:\mathrm{sin}\:\theta=−\frac{\mathrm{m}}{\mathrm{n}},\:\mathrm{where}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n} \\ $$$$\mathrm{prime}\:\mathrm{relatif}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{m}+\mathrm{n}\:\mathrm{is}\ldots \\…
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Question Number 174779 by mnjuly1970 last updated on 10/Aug/22 Commented by behi834171 last updated on 10/Aug/22 $$\boldsymbol{{sir}}!\:\:\boldsymbol{{your}}\:\boldsymbol{{used}}\:\boldsymbol{{fonts}}\:\boldsymbol{{are}}\:\boldsymbol{{too}}\:\boldsymbol{{small}}. \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{be}}\:\boldsymbol{{friend}}\:\boldsymbol{{with}}\:\boldsymbol{{my}}\:\boldsymbol{{eyes}}\:\boldsymbol{{and}} \\ $$$$\boldsymbol{{use}}\:\boldsymbol{{larger}}\:\boldsymbol{{fonts}}.\boldsymbol{{thanks}}\:\boldsymbol{{in}}\:\boldsymbol{{advance}}. \\ $$ Commented by…
Question Number 174666 by mnjuly1970 last updated on 07/Aug/22 Answered by behi834171 last updated on 08/Aug/22 $$\boldsymbol{{cos}}\frac{\boldsymbol{{C}}}{\mathrm{2}}=\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\boldsymbol{{ab}}}} \\ $$$$\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{\Psi}=\mathrm{1}−\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\Psi}=\mathrm{1}−\frac{\mathrm{4}\boldsymbol{{ab}}}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} }.\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\boldsymbol{{ab}}}= \\ $$$$=\mathrm{1}−\frac{\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)\left(\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{c}}\right)}{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} }=…