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Question Number 190151 by Matica last updated on 28/Mar/23

I saw this in a book (without explanation). Please show how. It is given that tan 2θ=(B/(A−C)) (A,B,C ∈R) . Find cos 2θ.

$${I}\:{saw}\:{this}\:{in}\:{a}\:{book}\:\left({without}\:{explanation}\right).\:{Please}\:{show}\:{how}. \\ $$$${It}\:{is}\:{given}\:{that}\:\mathrm{tan}\:\mathrm{2}\theta=\frac{{B}}{{A}−{C}}\:\:\left({A},{B},{C}\:\in\mathbb{R}\right)\:.\:{Find}\:\mathrm{cos}\:\mathrm{2}\theta. \\ $$

Commented by mr W last updated on 28/Mar/23

i saw this in an other book (without explanation, because nobody needs). tan α is given, how to find cos α?

$${i}\:{saw}\:{this}\:{in}\:{an}\:{other}\:{book}\:\left({without}\:\right. \\ $$$$\left.{explanation},\:{because}\:{nobody}\:{needs}\right).\: \\ $$$$\mathrm{tan}\:\alpha\:{is}\:{given},\:{how}\:{to}\:{find}\:\mathrm{cos}\:\alpha? \\ $$

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 cos 2θ without finding the value of θ . That makes me wonder.

$${in}\:{that}\:{book}\:,\:{the}\:{author}\:{calculate}\:\mathrm{cos}\:\mathrm{2}\theta\:{without}\:{finding}\:{the}\:{value}\:{of}\:\theta\:.\:{That}\:{makes}\:{me}\:{wonder}. \\ $$

Commented by Matica last updated on 30/Mar/23

thank you

$${thank}\:{you} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 28/Mar/23

tan^2 x+1=sec^2 x=(1/(cos^2 x)) ⇒cosx=±(1/( (√(tan^2 x+1)))) ⇒cos2θ=±(1/( (√(tan^2 2θ+1)))) =±((∣A−C∣)/( (√(A^2 +B^2 +C^2 −2AC))))

$$\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{1}=\mathrm{sec}^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$$\Rightarrow\mathrm{cos}{x}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{1}}} \\ $$$$\Rightarrow\mathrm{cos2}\theta=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \mathrm{2}\theta+\mathrm{1}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\pm\frac{\mid{A}−{C}\mid}{\:\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{C}^{\mathrm{2}} −\mathrm{2}{AC}}} \\ $$

Commented by Matica last updated on 30/Mar/23

Oh thank you

$${Oh}\:{thank}\:{you} \\ $$

Answered by manxsol last updated on 29/Mar/23

((sin^2 2θ)/(cos^2 2θ))=(B^2 /((A−C)^2 )) {+1} ((sin^2 2θ+cos^2 2θ)/(cos^2 2θ))=((B^2 +(A−C)^2 )/((A−C)^2 )) (1/(cos^2 2θ))=((B^2 +(A−C)^2 )/((A−C)^2 )) cos2θ=±((∣A−C∣)/( (√(B^2 +(A−C)^2 ))))

$$\frac{{sin}^{\mathrm{2}} \mathrm{2}\theta}{{cos}^{\mathrm{2}} \mathrm{2}\theta}=\frac{{B}^{\mathrm{2}} }{\left({A}−{C}\right)^{\mathrm{2}} }\:\:\:\:\:\:\:\:\left\{+\mathrm{1}\right\} \\ $$$$\frac{{sin}^{\mathrm{2}} \mathrm{2}\theta+{cos}^{\mathrm{2}} \mathrm{2}\theta}{{cos}^{\mathrm{2}} \mathrm{2}\theta}=\frac{{B}^{\mathrm{2}} +\left({A}−{C}\right)^{\mathrm{2}} }{\left({A}−{C}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{2}\theta}=\frac{{B}^{\mathrm{2}} +\left({A}−{C}\right)^{\mathrm{2}} }{\left({A}−{C}\right)^{\mathrm{2}} } \\ $$$${cos}\mathrm{2}\theta=\pm\frac{\mid{A}−{C}\mid}{\:\sqrt{{B}^{\mathrm{2}} +\left({A}−{C}\right)^{\mathrm{2}} }} \\ $$

Commented by Matica last updated on 30/Mar/23

thank you

$${thank}\:{you} \\ $$

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