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if-tanA-tanB-P-and-AtanB-q-express-the-value-of-cos-2A-3B-in-terms-of-p-and-q-

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Question Number 33969 by mondodotto@gmail.com last updated on 28/Apr/18
if tanA+tanB=P and AtanB=q express the value of cos(2A+3B) in terms of p and q.
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{tanA}}+\boldsymbol{\mathrm{tanB}}=\boldsymbol{\mathrm{P}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{AtanB}}=\boldsymbol{\mathrm{q}} \\ $$$$\boldsymbol{\mathrm{express}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{cos}}\left(\mathrm{2}\boldsymbol{\mathrm{A}}+\mathrm{3}\boldsymbol{\mathrm{B}}\right)\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{p}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{q}}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Apr/18
let tanA=a tanB=b cos(2A+2B+B) =cos(2A+2B).cosB−sin(2A+2B).sinB ={cos2A.cos2B−sin2A.sin2B}cosB−{sin2A. cos2B+cos2A.sin2B}.sinB put cos2A= (1−a^2 /1+a^2 ) sin2A=2a/1+a^2 simplyfy if prlblem comment pls..
$${let}\:{tanA}={a}\:\:{tanB}={b}\: \\ $$$${cos}\left(\mathrm{2}{A}+\mathrm{2}{B}+{B}\right) \\ $$$$={cos}\left(\mathrm{2}{A}+\mathrm{2}{B}\right).{cosB}−{sin}\left(\mathrm{2}{A}+\mathrm{2}{B}\right).{sinB} \\ $$$$=\left\{{cos}\mathrm{2}{A}.{cos}\mathrm{2}{B}−{sin}\mathrm{2}{A}.{sin}\mathrm{2}{B}\right\}{cosB}−\left\{{sin}\mathrm{2}{A}.\right. \\ $$$$\left.{cos}\mathrm{2}{B}+{cos}\mathrm{2}{A}.{sin}\mathrm{2}{B}\right\}.{sinB} \\ $$$${put}\:{cos}\mathrm{2}{A}=\:\left(\mathrm{1}−{a}^{\mathrm{2}} /\mathrm{1}+{a}^{\mathrm{2}} \right)\:{sin}\mathrm{2}{A}=\mathrm{2}{a}/\mathrm{1}+{a}^{\mathrm{2}} \\ $$$${simplyfy}\:{if}\:{prlblem}\:{comment}\:{pls}.. \\ $$$$ \\ $$
Commented by mondodotto@gmail.com last updated on 29/Apr/18
i dont undstnd
$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{dont}}\:\boldsymbol{\mathrm{undstnd}} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Apr/18
ok i am solving in details and shall upload the image
$${ok}\:{i}\:{am}\:{solving}\:{in}\:{details}\:{and}\:{shall}\:{upload}\:{the}\:{image} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Apr/18
let tanA=a,tanB=b so a+b=p ,ab=q (a−b)^2 =(a+b)^2 −4ab a−b=(√(p^2 −4q)) and a+b=p solving a=p+(√(p^2 −4q )) b=p−(√(p^2 −4q)) cos(2A+3B) =cos2A.cos3B−sin2A.sin3B =(((1−tan^2 A)/(1+tan^2 A)))cosB(3−4cos^2 B)−(((2tanA)/(1+tan^2 A))).sinB. (4sin^2 B−3) now replace tanA by a tanB by b cosB=(1/( (√(1+b^2 )) )) sinB=((tanB)/(secB))=(b/( (√(1+b^2 )) )) ((1−a^2 )/(1+a^2 )).(1/( (√(1+b^2 ))))(3−(4/(1+b^2 )))−(((2a)/(1+a^2 )))(b/( (√(1+b^2 )))).(((4b^2 )/(1+b^2 ))−3 put a=p+(√(p^2 −4q )) and b=p−(√(p^2 4q)) simplify
$${let}\:{tanA}={a},{tanB}={b}\:{so}\:{a}+{b}={p}\:,{ab}={q} \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} \:−\mathrm{4}{ab} \\ $$$${a}−{b}=\sqrt{{p}^{\mathrm{2}} −\mathrm{4}{q}}\:\:{and}\:\:{a}+{b}={p} \\ $$$${solving}\:{a}={p}+\sqrt{{p}^{\mathrm{2}} \:−\mathrm{4}{q}\:} \\ $$$${b}={p}−\sqrt{{p}^{\mathrm{2}} −\mathrm{4}{q}}\:\: \\ $$$${cos}\left(\mathrm{2}{A}+\mathrm{3}{B}\right) \\ $$$$={cos}\mathrm{2}{A}.{cos}\mathrm{3}{B}−{sin}\mathrm{2}{A}.{sin}\mathrm{3}{B} \\ $$$$=\left(\frac{\mathrm{1}−{tan}^{\mathrm{2}} {A}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}\right){cosB}\left(\mathrm{3}−\mathrm{4}{cos}^{\mathrm{2}} {B}\right)−\left(\frac{\mathrm{2}{tanA}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}\right).{sinB}. \\ $$$$\left(\mathrm{4}{sin}^{\mathrm{2}} {B}−\mathrm{3}\right) \\ $$$${now}\:{replace}\:{tanA}\:{by}\:{a}\:\:{tanB}\:{by}\:{b} \\ $$$${cosB}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }\:\:}\:\:{sinB}=\frac{{tanB}}{{secB}}=\frac{{b}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }\:\:}\: \\ $$$$\frac{\mathrm{1}−{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} }.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}\left(\mathrm{3}−\frac{\mathrm{4}}{\mathrm{1}+{b}^{\mathrm{2}} }\right)−\left(\frac{\mathrm{2}{a}}{\mathrm{1}+{a}^{\mathrm{2}} }\right)\frac{{b}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}.\left(\frac{\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} }−\mathrm{3}\right. \\ $$$${put}\:{a}={p}+\sqrt{{p}^{\mathrm{2}} \:\:−\mathrm{4}{q}\:}\:{and}\:{b}={p}−\sqrt{{p}^{\mathrm{2}} \:\mathrm{4}{q}} \\ $$$${simplify} \\ $$$$ \\ $$

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