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I-saw-this-in-a-book-without-explanation-Please-show-how-It-is-given-that-tan-2-B-A-C-A-B-C-R-Find-cos-2-




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.
Isawthisinabook(withoutexplanation).Pleaseshowhow.Itisgiventhattan2θ=BAC(A,B,CR).Findcos2θ.
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?
isawthisinanotherbook(withoutexplanation,becausenobodyneeds).tanαisgiven,howtofindcosα?
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.
inthatbook,theauthorcalculatecos2θwithoutfindingthevalueofθ.Thatmakesmewonder.
Commented by Matica last updated on 30/Mar/23
thank you
thankyou
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))))
tan2x+1=sec2x=1cos2xcosx=±1tan2x+1cos2θ=±1tan22θ+1=±ACA2+B2+C22AC
Commented by Matica last updated on 30/Mar/23
Oh thank you
Ohthankyou
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 ))))
sin22θcos22θ=B2(AC)2{+1}sin22θ+cos22θcos22θ=B2+(AC)2(AC)21cos22θ=B2+(AC)2(AC)2cos2θ=±ACB2+(AC)2
Commented by Matica last updated on 30/Mar/23
thank you
thankyou

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