Question Number 137811 by bramlexs22 last updated on 07/Apr/21 $${Given}\:\begin{cases}{\mathrm{cos}\:\left({x}−{y}\right)=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{x}}\\{\mathrm{cos}\:\left({x}+{y}\right)=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:{x}}\end{cases} \\ $$$${where}\:\mathrm{270}°\:<\:{y}<\:\mathrm{360}°\:.\:{find} \\ $$$$\mathrm{sin}\:\mathrm{2}{y}\:. \\ $$ Answered by EDWIN88 last updated on 07/Apr/21 $$\Leftrightarrow\:\mathrm{cos}\:\left({x}−{y}\right)+\mathrm{cos}\:\left({x}+{y}\right)=\frac{\mathrm{5}}{\mathrm{6}}\mathrm{cos}\:{x} \\…
Question Number 6736 by Tawakalitu. last updated on 18/Jul/16 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 6723 by Tawakalitu. last updated on 16/Jul/16 $${An}\:{MTN}\:{mask}\:{is}\:{erected}\:{at}\:{a}\:{point}\:{P}\:\:{in}\:{ilaro}\:{town}.\:{At}\:{a}\:{point} \\ $$$${B}\:{due}\:{west},\:{The}\:{angle}\:{of}\:{elevation}\:{of}\:{its}\:{top}\:{is}\:\beta\:{and}\:{at}\:{point}\: \\ $$$${C}\:{due}\:{south},\:{the}\:{angle}\:{of}\:{elevation}\:{is}\:\alpha.\:{With}\:{the}\:{aid}\:{of}\:{an}\: \\ $$$${appropriate}\:{diagam}.\:{show}\:{that}\:{the}\:{angle}\:{of}\:{elevation}\:{of}\:{the}\:{top} \\ $$$${from}\:{a}\:{point}\:{due}\:{south}\:{of}\:{B}\:{and}\:{due}\:{west}\:{of}\:{C}\:{is}\: \\ $$$$ \\ $$$${cot}^{−\mathrm{1}} \left[{cot}^{\mathrm{2}} \left(\beta\right)\:+\:{cot}^{\mathrm{2}} \left(\alpha\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 6715 by Tawakalitu. last updated on 15/Jul/16 $${Solve}\:{the}\:{following}\:{equation}\:{for}\:\mathrm{0}\:<\:\Theta\:<\:\mathrm{360}^{{o}} \\ $$$${cosx}\:+\:{cos}\mathrm{3}{x}\:+\:{cos}\mathrm{5}{x}\:+\:{cos}\mathrm{7}{x}\:=\:\mathrm{0} \\ $$ Answered by Yozzii last updated on 16/Jul/16 $${We}\:{know}\:{the}\:{identity}\:{cosa}+{cosb}=\mathrm{2}{cos}\frac{{a}+{b}}{\mathrm{2}}{cos}\frac{{a}−{b}}{\mathrm{2}}\:{for}\:{a},{b}\in\mathbb{R}….\left(\mathrm{1}\right) \\ $$$${Let}\:{u}={cosx}+{cos}\mathrm{3}{x}+{cos}\mathrm{5}{x}+{cos}\mathrm{7}{x}. \\…
Question Number 137771 by bemath last updated on 06/Apr/21 Commented by bemath last updated on 06/Apr/21 $${any}\:{one}\:{solve}\:{the}\:{question}\:{number} \\ $$$$\mathrm{16}? \\ $$ Commented by Ñï= last…
Question Number 6655 by Tawakalitu. last updated on 09/Jul/16 $${If}\:\:\:{tan}\mathrm{2}{x}\:−\:{sin}\mathrm{2}{x}\:=\:{b}\:\:{and}\:\:{tan}\mathrm{2}{x}\:+\:{sin}\mathrm{2}{x}\:=\:{a} \\ $$$${prove}\:{that}\::\: \\ $$$${b}^{\mathrm{2}} \:−\:{a}^{\mathrm{2}\:} \:=\:\mathrm{16}{ba} \\ $$ Commented by Yozzii last updated on 09/Jul/16…
Question Number 6602 by cindycastro last updated on 05/Jul/16 $$\%{ml} \\ $$ Commented by Temp last updated on 05/Jul/16 $$=\frac{{ml}}{\mathrm{100}} \\ $$ Terms of Service…
Question Number 6558 by Tawakalitu. last updated on 02/Jul/16 $${If}\:\:{secx}\:−\:{tanx}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${find}\:\:\:{secx}\:+\:{tanx} \\ $$ Commented by Tawakalitu. last updated on 02/Jul/16 $$ \\ $$$$ \\…
Question Number 71918 by lalitchand last updated on 22/Oct/19 $$\mathrm{If}\:\mathrm{Cos}\theta=\frac{\mathrm{x}\:\mathrm{cos}\beta\:−\:\mathrm{y}}{\mathrm{x}\:−\:\mathrm{y}\:\mathrm{cos}\beta}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\:\mathrm{tan}\frac{\theta}{\mathrm{2}}\:=\sqrt{\frac{\mathrm{x}−\mathrm{y}}{\mathrm{x}+\mathrm{y}}\:}\:\mathrm{tan}\frac{\beta}{\mathrm{2}} \\ $$ Commented by Prithwish sen last updated on 22/Oct/19 $$\mathrm{Using}\:\mathrm{componendo}\:\mathrm{dividendo} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}+\mathrm{cos}\theta}\:=\:\frac{\left(\mathrm{1}−\mathrm{cos}\beta\right)\left(\mathrm{x}+\mathrm{y}\right)}{\left(\mathrm{1}+\mathrm{cos}\beta\right)\left(\mathrm{x}−\mathrm{y}\right)} \\ $$$$\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\theta}}{\mathrm{2}}\:=\:\sqrt{\frac{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}}}\:\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\beta}}{\mathrm{2}}…
Question Number 6370 by sanusihammed last updated on 25/Jun/16 $${If}\:\:{x}\:=\:\frac{\mathrm{3}{sin}\theta}{\:\sqrt{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:}}\:\:\:{and}\:\:{y}\:=\:\mathrm{4}{cos}\theta \\ $$$${find}\:\:{cos}\theta \\ $$ Commented by nburiburu last updated on 25/Jun/16 $${cos}\theta={y}/\mathrm{4}\:{with}\:\mathrm{16}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}}…