Question Number 190151 by Matica last updated on 28/Mar/23
$${I}\:{saw}\:{this}\:{in}\:{a}\:{book}\:\left({without}\:{explanation}\right).\:{Please}\:{show}\:{how}. \\ $$$${It}\:{is}\:{given}\:{that}\:\mathrm{tan}\:\mathrm{2}\theta=\frac{{B}}{{A}−{C}}\:\:\left({A},{B},{C}\:\in\mathbb{R}\right)\:.\:{Find}\:\mathrm{cos}\:\mathrm{2}\theta. \\ $$
Commented by mr W last updated on 28/Mar/23
$${i}\:{saw}\:{this}\:{in}\:{an}\:{other}\:{book}\:\left({without}\:\right. \\ $$$$\left.{explanation},\:{because}\:{nobody}\:{needs}\right).\: \\ $$$$\mathrm{tan}\:\alpha\:{is}\:{given},\:{how}\:{to}\:{find}\:\mathrm{cos}\:\alpha? \\ $$
Commented by mr W last updated on 28/Mar/23
Commented by Matica last updated on 30/Mar/23
$${in}\:{that}\:{book}\:,\:{the}\:{author}\:{calculate}\:\mathrm{cos}\:\mathrm{2}\theta\:{without}\:{finding}\:{the}\:{value}\:{of}\:\theta\:.\:{That}\:{makes}\:{me}\:{wonder}. \\ $$
Commented by Matica last updated on 30/Mar/23
$${thank}\:{you} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 28/Mar/23
$$\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{1}=\mathrm{sec}^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$$\Rightarrow\mathrm{cos}{x}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{1}}} \\ $$$$\Rightarrow\mathrm{cos2}\theta=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \mathrm{2}\theta+\mathrm{1}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\pm\frac{\mid{A}−{C}\mid}{\:\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{C}^{\mathrm{2}} −\mathrm{2}{AC}}} \\ $$
Commented by Matica last updated on 30/Mar/23
$${Oh}\:{thank}\:{you} \\ $$
Answered by manxsol last updated on 29/Mar/23
$$\frac{{sin}^{\mathrm{2}} \mathrm{2}\theta}{{cos}^{\mathrm{2}} \mathrm{2}\theta}=\frac{{B}^{\mathrm{2}} }{\left({A}−{C}\right)^{\mathrm{2}} }\:\:\:\:\:\:\:\:\left\{+\mathrm{1}\right\} \\ $$$$\frac{{sin}^{\mathrm{2}} \mathrm{2}\theta+{cos}^{\mathrm{2}} \mathrm{2}\theta}{{cos}^{\mathrm{2}} \mathrm{2}\theta}=\frac{{B}^{\mathrm{2}} +\left({A}−{C}\right)^{\mathrm{2}} }{\left({A}−{C}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{2}\theta}=\frac{{B}^{\mathrm{2}} +\left({A}−{C}\right)^{\mathrm{2}} }{\left({A}−{C}\right)^{\mathrm{2}} } \\ $$$${cos}\mathrm{2}\theta=\pm\frac{\mid{A}−{C}\mid}{\:\sqrt{{B}^{\mathrm{2}} +\left({A}−{C}\right)^{\mathrm{2}} }} \\ $$
Commented by Matica last updated on 30/Mar/23
$${thank}\:{you} \\ $$