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Question Number 54126 by rahul 19 last updated on 29/Jan/19 $${Prove}\:{that} \\ $$$$\frac{\mathrm{sin}\:\mathrm{19}\theta}{\mathrm{sin}\:\theta}\:=\:\mathrm{cos}\left(−\mathrm{18}\theta\right)+\mathrm{cos}\left(−\mathrm{16}\theta\right)+… \\ $$$$\:\:\:…+\:\mathrm{cos}\left(−\mathrm{2}\theta\right)+\mathrm{1}+\mathrm{cos}\left(\mathrm{2}\theta\right)+….+\mathrm{cos}\left(\mathrm{18}\theta\right). \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19 $$…
Question Number 185134 by emmanuelson123 last updated on 17/Jan/23 Answered by som(math1967) last updated on 17/Jan/23 $$\:\frac{{sin}^{\mathrm{4}} {x}}{\mathrm{5}}\:+\frac{\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow\mathrm{7}{sin}^{\mathrm{4}} {x}+\mathrm{5}−\mathrm{10}{sin}^{\mathrm{2}} {x}+\mathrm{5}{sin}^{\mathrm{4}} {x}=\frac{\mathrm{35}}{\mathrm{12}}…
Question Number 185107 by emmanuelson123 last updated on 17/Jan/23 Answered by SEKRET last updated on 17/Jan/23 $$\:\:\:\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2048}}\right)\:+\:\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{2048}}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{2}\left(\mathrm{1}−\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2048}}\right)\right)\:+\:\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2048}}\right)\:−\:\mathrm{1}=\mathrm{1} \\ $$$$\:\:\mathrm{2}−\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2048}}\right)+\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}}…
Question Number 185101 by emmanuelson123 last updated on 17/Jan/23 Answered by Frix last updated on 17/Jan/23 $$\mathrm{In}\:\mathrm{triangle}\:{a},\:{b},\:{c}: \\ $$$${D}=\sqrt{\left({a}+{b}+{c}\right)\left({b}+{c}−{a}\right)\left({a}+{c}−{b}\right)\left({a}+{b}−{c}\right)} \\ $$$$ \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\gamma\:={c}^{\mathrm{2}}…
Question Number 119548 by benjo_mathlover last updated on 25/Oct/20 Commented by benjo_mathlover last updated on 25/Oct/20 $$\:{If}\:\mathrm{cos}\:{A}\:=\:\mathrm{tan}\:{B}\:,\:\mathrm{cos}\:{B}\:=\:\mathrm{tan}\:{C}\:{and}\: \\ $$$$\mathrm{cos}\:{C}\:=\:\mathrm{tan}\:{A}\:,\:{then}\:\mathrm{sin}\:{A}\:{is}\:{equal}\:{to} \\ $$ Commented by TANMAY PANACEA…
Question Number 119506 by bemath last updated on 25/Oct/20 $$\:\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{5}}\mathrm{cos}\:{x}+\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\mathrm{sin}\:{x}\:\leqslant\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$ Answered by 1549442205PVT last updated on 25/Oct/20 $$\:\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{5}}\mathrm{cos}\:{x}+\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\mathrm{sin}\:{x}\:\leqslant\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{cos}\left(\mathrm{x}−\frac{\pi}{\mathrm{5}}\right)\leqslant\mathrm{cos}\frac{\pi}{\mathrm{4}} \\ $$$$\Leftrightarrow\mathrm{x}−\frac{\pi}{\mathrm{5}}\geqslant\frac{\pi}{\mathrm{4}}+\mathrm{2m}\pi \\…
Question Number 119494 by liberty last updated on 25/Oct/20 $${Russian}\:{olympiad}\: \\ $$$${find}\:{real}\:{solution}\:{of}\:{the}\:{system}\: \\ $$$$\begin{cases}{\mathrm{sin}\:{x}+\mathrm{2sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\\{\mathrm{sin}\:{y}+\mathrm{3sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\\{\mathrm{sin}\:{z}+\mathrm{4sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\end{cases} \\ $$$$ \\ $$ Answered by benjo_mathlover last updated on 25/Oct/20…
Question Number 119421 by bemath last updated on 24/Oct/20 $${find}\:{the}\:{value}\:{of}\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{999}} {\prod}}\:\mathrm{cos}\:\left({ia}\right)\:;\:{where}\:{a}\:=\:\frac{\mathrm{2}\pi}{\mathrm{1999}} \\ $$ Commented by MJS_new last updated on 24/Oct/20 $$\mathrm{study}\:{P}_{{n}} =\underset{{j}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\:\frac{\mathrm{2}\pi{j}}{\mathrm{2}{n}+\mathrm{1}}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{answer}…