Question Number 68964 by ahmadshah last updated on 17/Sep/19 Commented by mind is power last updated on 17/Sep/19 $${m}+{in}={e}^{{ia}} +{e}^{{ib}} \Rightarrow{m}^{\mathrm{2}} −{n}^{\mathrm{2}} +\mathrm{2}{imn}={e}^{{i}\mathrm{2}{a}} +{e}^{\mathrm{2}{ib}} +\mathrm{2}{e}^{{i}\left({a}+{b}\right)}…
Question Number 134479 by abdullahquwatan last updated on 04/Mar/21 $$\mathrm{sin}\:\mathrm{x}+\mathrm{sin}\:\mathrm{2x}+\mathrm{sin}\:\mathrm{3x}=\mathrm{cos}\:\mathrm{x}+\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{0}<\mathrm{x}<\pi \\ $$ Answered by EDWIN88 last updated on 14/Mar/21 $$\mathrm{sin}\:\mathrm{3x}+\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{2sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{x} \\ $$$$\left(\bullet\right)\mathrm{2sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{sin}\:\mathrm{2x}\:=\:\mathrm{cos}\:\mathrm{x}\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{x}\right)…
Question Number 134475 by EDWIN88 last updated on 04/Mar/21 $$\mathrm{If}\:\mathrm{1}+\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{2x}+\mathrm{cos}\:\mathrm{3x}+\mathrm{cos}\:\mathrm{4x}+…+\infty\:=\:\mathrm{3} \\ $$$$\mathrm{for}\:\mathrm{0}<\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:\mathrm{x}+\mathrm{sin}\:\mathrm{2x}+\mathrm{sin}\:\mathrm{3x}+…+\infty \\ $$ Commented by Dwaipayan Shikari last updated on 04/Mar/21 $$\underset{{n}=\mathrm{0}}…
Question Number 134464 by bramlexs22 last updated on 04/Mar/21 $$\:\:\:\:\:\:\:\begin{array}{|c|c|}{\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{Equation}}\\{\:\:\mathrm{81}^{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \:+\:\mathrm{81}^{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}} \:=\:\mathrm{30}\:}\\\hline\end{array} \\ $$ Commented by harckinwunmy last updated on 04/Mar/21 $${ff} \\…
Question Number 134461 by pticantor last updated on 04/Mar/21 $${soit}\:\left(\boldsymbol{{E}}\right)\:{l}'{equation}\:{complex}: \\ $$$$\boldsymbol{{z}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{pz}}−\boldsymbol{{q}}=\mathrm{0} \\ $$$$\boldsymbol{{p}},\boldsymbol{{q}}\in\mathbb{C} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}\in\mathbb{C}\backslash\:\boldsymbol{{u}}^{\mathrm{2}} =\boldsymbol{{q}} \\ $$$$\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{if}}\:\boldsymbol{{z}}_{\mathrm{1}} ,\boldsymbol{{z}}_{\mathrm{2}} \boldsymbol{{are}}\:\boldsymbol{{solutions}}\:\boldsymbol{{of}}\:\left(\boldsymbol{{E}}\right), \\ $$$$\mid\boldsymbol{{z}}_{\mathrm{1}} \mid+\mid\boldsymbol{{z}}_{\mathrm{2}}…
Question Number 134446 by physicstutes last updated on 03/Mar/21 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\:\frac{\mathrm{cos}^{\mathrm{4}} {x}\:+\:\mathrm{sin}^{\mathrm{4}} {x}}{\mathrm{cos}^{\mathrm{4}} {x}−\mathrm{sin}^{\mathrm{4}} {x}}\:=\:\frac{\mathrm{cos2}{A}\:+\:\mathrm{sec}\:\mathrm{2}{A}}{\mathrm{2}} \\ $$ Answered by Ar Brandon last updated on…
Question Number 134417 by faysal last updated on 03/Mar/21 Answered by som(math1967) last updated on 03/Mar/21 $$\frac{\mathrm{2}^{{n}−\mathrm{1}} \mathrm{2}{sin}\theta{cos}\theta.{cos}\mathrm{2}\theta…{cos}\mathrm{2}^{{n}−\mathrm{1}} \theta}{{sin}\theta} \\ $$$$\frac{\mathrm{2}^{{n}−\mathrm{2}} .\mathrm{2}{sin}\mathrm{2}\theta{cos}\mathrm{2}\theta.{cos}\mathrm{2}^{\mathrm{2}} \theta…..}{{sin}\theta} \\ $$$$\frac{\mathrm{2}^{{n}−\mathrm{3}}…
Question Number 3317 by rishabh last updated on 10/Dec/15 $$\mathrm{If}\:\mathrm{cosec}\:\theta,\:\mathrm{sec}\:\theta,\:\mathrm{and}\:\mathrm{cot}\:\theta\:\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}.,\:\mathrm{then} \\ $$$$\frac{\mathrm{sin}\:\theta+\mathrm{tan}\:\theta}{\mathrm{cos}\:\theta}\:=\:? \\ $$ Answered by prakash jain last updated on 10/Dec/15 $$\frac{\mathrm{2}}{\mathrm{sec}\:\theta}=\frac{\mathrm{1}}{\mathrm{cosec}\:\theta}+\frac{\mathrm{1}}{\mathrm{cot}\:\theta}=\mathrm{sin}\:\theta+\mathrm{tan}\:\theta \\ $$$$\mathrm{2cos}\:\theta=\mathrm{sin}\:\theta+\mathrm{tan}\:\theta…
Question Number 68761 by Rasheed.Sindhi last updated on 15/Sep/19 $$\mathrm{Two}\:\boldsymbol{\mathrm{arcs}}\:\mathrm{having}\:\mathrm{their}\:\mathrm{centers}\:\mathrm{on}\:\mathrm{a} \\ $$$$\boldsymbol{\mathrm{circle}}\:\mathrm{are}\:\mathrm{cutting}\:\mathrm{each}\:\mathrm{other}\:\mathrm{at}\:\mathrm{a}\: \\ $$$$\mathrm{single}\:\mathrm{point}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{and}\:\mathrm{thus} \\ $$$$\:\mathrm{dividing}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{four}\:\mathrm{regions}. \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{arcs}\:\mathrm{cut}\:\mathrm{each}\:\mathrm{other}\:\mathrm{in}\:\boldsymbol{\mathrm{a}}:\boldsymbol{\mathrm{b}}\:\&\:\boldsymbol{\mathrm{c}}:\boldsymbol{\mathrm{d}}\: \\ $$$$\mathrm{ratios}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{between}\:\mathrm{four} \\ $$$$\mathrm{regions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{when}\:\mathrm{the}\:\mathrm{circle} \\…
Question Number 68673 by mr W last updated on 14/Sep/19 $${find}\:\boldsymbol{\mathrm{sin}}\:\mathrm{20}°=? \\ $$ Commented by MJS last updated on 15/Sep/19 $$\Rightarrow\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{use}\:\mathrm{this}\:\mathrm{to}\:\mathrm{get}\:\mathrm{sin}\:\mathrm{1}°,\:\mathrm{because} \\ $$$$\mathrm{sin}\:\mathrm{3}°\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}−\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{2}} \\ $$$$\mathrm{and}…