Question Number 119109 by bagjagunawan last updated on 22/Oct/20 Answered by 1549442205PVT last updated on 22/Oct/20 $$\mathrm{P}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{9}\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{27}\pi}{\mathrm{20}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos9}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos81}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos27}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos63}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos9}+\mathrm{cos81}}{\mathrm{2}}+\mathrm{cos9cos81}\right) \\ $$$$×\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos27}−\mathrm{cos63}}{\mathrm{2}}−\mathrm{cos27cos63}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{cos45cos36}+\frac{\mathrm{cos90}+\mathrm{cos72}}{\mathrm{2}}\right)…
Question Number 119015 by benjo_mathlover last updated on 21/Oct/20 $${x}\:=\:\frac{\mathrm{sin}\:\mathrm{1}°+\mathrm{sin}\:\mathrm{2}°+\mathrm{sin}\:\mathrm{3}°+…+\mathrm{sin}\:\mathrm{45}°}{\mathrm{cos}\:\mathrm{1}°+\mathrm{cos}\:\mathrm{2}°+\mathrm{cos}\:\mathrm{3}°+…+\mathrm{cos}\:\mathrm{45}°} \\ $$$${x}\:=\:? \\ $$ Commented by Dwaipayan Shikari last updated on 21/Oct/20 $${tan}\left(\mathrm{1}°+\frac{\mathrm{45}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)={tan}\mathrm{23}° \\ $$…
Question Number 119007 by bramlexs22 last updated on 21/Oct/20 $${Given}\:{point}\:{A}=\left(\mathrm{7},\mathrm{26}\right)\:{and}\:{B}=\left(\mathrm{12},\mathrm{12}\right)\:,\:{find} \\ $$$${all}\:{points}\:{P}\:{such}\:{that}\:\mid{AP}\:\mid\:=\:\mid\:{BP}\:\mid\:{and}\:\angle{APB}\:=\:\mathrm{90}°\:. \\ $$ Answered by bemath last updated on 21/Oct/20 $$\:{Note}\:{that}\:{the}\:{triangle}\:{ABP}\:{is}\:{an}\:{isosceles}\:{right} \\ $$$${triangle}\:{with}\:\mid{AP}\:\mid\:=\:\mid{BP}\:\mid.\:{Let}\:{Q}\:{be}\:{the}\:{midpoint} \\…
Question Number 184538 by Kiuanush last updated on 08/Jan/23 $$ \\ $$ Commented by Frix last updated on 08/Jan/23 $$\: \\ $$ Answered by Frix…
Question Number 53447 by rajeshghorai130@gmail.com last updated on 22/Jan/19 Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19 $${a}={sin}^{\mathrm{2}} {x} \\ $$$${b}={sin}^{\mathrm{2}} {y} \\ $$$$\frac{\left(\mathrm{1}−{a}\right)^{\mathrm{2}} }{\mathrm{1}−{b}}+\frac{{a}^{\mathrm{2}} }{{b}}=\mathrm{1}…
Question Number 118899 by bobhans last updated on 20/Oct/20 $$\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{14}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{14}}\right)…\mathrm{sin}\:\left(\frac{\mathrm{6}\pi}{\mathrm{14}}\right)=? \\ $$ Answered by TANMAY PANACEA last updated on 20/Oct/20 $${taking}\:{help}\:{from}\:{S}.{L}\:{Loney}\:{Trigonometry} \\ $$$$\mathrm{2}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {sin}\left(\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\right){sin}\left(\frac{\mathrm{4}\pi}{\mathrm{2}{n}}\right){sin}\left(\frac{\mathrm{6}\pi}{\mathrm{2}{n}}\right)…{sin}\left(\frac{{n}−\mathrm{2}}{\mathrm{2}{n}}\right)\pi=\sqrt{{n}}\: \\…
Question Number 53340 by aseerimad last updated on 20/Jan/19 Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19 $${a}=\mathrm{18}^{{o}} \\ $$$$ \\ $$$${method}\:\mathrm{1} \\ $$$$\frac{{sin}\mathrm{54}}{{cos}\mathrm{54}}−\frac{{sin}\mathrm{36}}{{cos}\mathrm{36}} \\ $$$$\frac{{sin}\mathrm{54}{cos}\mathrm{36}−{cos}\mathrm{54}{sin}\mathrm{36}}{{cos}\mathrm{54}{cos}\mathrm{36}}…
Question Number 118861 by pqr last updated on 20/Oct/20 $${sin}^{−\mathrm{1}} \left(−\mathrm{1}/\mathrm{2}\right) \\ $$ Answered by MJS_new last updated on 20/Oct/20 $$−\frac{\pi}{\mathrm{6}} \\ $$ Terms of…
Question Number 53210 by Tawa1 last updated on 19/Jan/19 Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19 $${r}_{\mathrm{1}} ^{'} ={r}_{\mathrm{1}} −{r}_{\mathrm{3}} \\ $$$${r}_{\mathrm{2}} ^{,} ={r}_{\mathrm{2}} −{r}_{\mathrm{3}}…
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