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Question Number 199771 by Rupesh123 last updated on 09/Nov/23

Answered by deleteduser1 last updated on 09/Nov/23

((sin(4x)=2sin(2x)cos(2x))/a)=((sin(2x))/b) ⇒cos(2x)=(a/(2b))=2cos^2 x−1⇒cos^2 x=((a+2b)/(4b)) ((sin(2x)=2sin(x)cos(x))/b)=((sin(x))/c)⇒cos(x)=(b/(2c)) ⇒((a+2b)/b)=(b^2 /c^2 )⇒(a/b)=(b^2 /c^2 )−2

$$\frac{{sin}\left(\mathrm{4}{x}\right)=\mathrm{2}{sin}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{2}{x}\right)}{{a}}=\frac{{sin}\left(\mathrm{2}{x}\right)}{{b}} \\ $$$$\Rightarrow{cos}\left(\mathrm{2}{x}\right)=\frac{{a}}{\mathrm{2}{b}}=\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\Rightarrow{cos}^{\mathrm{2}} {x}=\frac{{a}+\mathrm{2}{b}}{\mathrm{4}{b}} \\ $$$$\frac{{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)}{{b}}=\frac{{sin}\left({x}\right)}{{c}}\Rightarrow{cos}\left({x}\right)=\frac{{b}}{\mathrm{2}{c}} \\ $$$$\Rightarrow\frac{{a}+\mathrm{2}{b}}{{b}}=\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\Rightarrow\frac{{a}}{{b}}=\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }−\mathrm{2} \\ $$

Commented by Mingma last updated on 09/Nov/23

Nice, sir!

Answered by Rasheed.Sindhi last updated on 09/Nov/23

(a/(sin 4θ))=(b/(sin 2θ))=(c/(sin θ))=k (say) a=k sin 4θ,b=k sin 2θ,c=k sin θ (a/b)=((b^2 −c^2 )/(a^2 −b^2 ))⇒ ((k sin 4θ)/(k sin 2θ))=((k^2 sin^2 2θ−k^2 sin^2 θ )/(k^2 sin^2 4θ−k^2 sin^2 2θ)) ((sin 4θ)/( sin 2θ))=((sin^2 2θ−sin^2 θ )/(sin^2 4θ−sin^2 2θ)) lhs: ((sin 4θ)/( sin 2θ))=((2 sin 2θ cos 2θ)/(sin 2θ))=2 cos 2θ rhs:((sin^2 2θ−sin^2 θ )/(sin^2 4θ−sin^2 2θ)) ....

$$\frac{{a}}{\mathrm{sin}\:\mathrm{4}\theta}=\frac{{b}}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{{c}}{\mathrm{sin}\:\theta}={k}\:\left({say}\right) \\ $$$${a}={k}\:\mathrm{sin}\:\mathrm{4}\theta,{b}={k}\:\mathrm{sin}\:\mathrm{2}\theta,{c}={k}\:\mathrm{sin}\:\theta \\ $$$$\frac{{a}}{{b}}=\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\Rightarrow \\ $$$$\:\:\:\:\frac{{k}\:\mathrm{sin}\:\mathrm{4}\theta}{{k}\:\mathrm{sin}\:\mathrm{2}\theta}=\frac{{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta\:}{{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{4}\theta−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$\:\:\:\:\frac{\mathrm{sin}\:\mathrm{4}\theta}{\:\mathrm{sin}\:\mathrm{2}\theta}=\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{sin}^{\mathrm{2}} \theta\:}{\mathrm{sin}^{\mathrm{2}} \mathrm{4}\theta−\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$${lhs}:\:\:\frac{\mathrm{sin}\:\mathrm{4}\theta}{\:\mathrm{sin}\:\mathrm{2}\theta}=\frac{\mathrm{2}\:\cancel{\mathrm{sin}\:\mathrm{2}\theta}\:\mathrm{cos}\:\mathrm{2}\theta}{\cancel{\mathrm{sin}\:\mathrm{2}\theta}}=\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$${rhs}:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{sin}^{\mathrm{2}} \theta\:}{\mathrm{sin}^{\mathrm{2}} \mathrm{4}\theta−\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$.... \\ $$

Commented by Rasheed.Sindhi last updated on 09/Nov/23

But sir since sin((π/7)) has no “nice” value, so there′s no “nice solution” of the question?

$${But}\:{sir}\:{since}\:\mathrm{sin}\left(\frac{\pi}{\mathrm{7}}\right)\:{has}\:{no}\:``{nice}'' \\ $$$${value},\:{so}\:{there}'{s}\:{no}\:``{nice}\:{solution}'' \\ $$$${of}\:{the}\:{question}? \\ $$

Commented by Frix last updated on 09/Nov/23

We know that (4+2+1)θ=π ⇒ θ=(π/7) Should be easy to show...

$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\left(\mathrm{4}+\mathrm{2}+\mathrm{1}\right)\theta=\pi\:\Rightarrow\:\theta=\frac{\pi}{\mathrm{7}} \\ $$$$\mathrm{Should}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{show}... \\ $$

Commented by Rasheed.Sindhi last updated on 09/Nov/23

Thanks dear sir for useful hint.

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\mathrm{dear}\:\boldsymbol{\mathrm{sir}}\:\mathrm{for}\:\mathrm{useful}\:\mathrm{hint}. \\ $$

Commented by Frix last updated on 09/Nov/23

I found a way, see my answer.

$$\mathrm{I}\:\mathrm{found}\:\mathrm{a}\:\mathrm{way},\:\mathrm{see}\:\mathrm{my}\:\mathrm{answer}. \\ $$

Commented by Mingma last updated on 09/Nov/23

Nice, sir!

Answered by Frix last updated on 09/Nov/23

(a/b)=((b^2 −c^2 )/(a^2 −b^2 )) ⇔ c^2 =−(a^3 /b)+ab+b^2 We know (a/(sin 4θ))=(b/(sin 2θ))=(c/(sin θ)) and 4θ+2θ+θ=π ⇒ θ=(π/7) Let x=2θ=((2π)/7) Let a=sin 2x ∧b=sin x ∧c=sin (x/2) ⇔ a=2cos x sin x ∧b=sin x ∧c^2 =((1−cos x)/2) Inserting we get ((1−cos x)/2)=−8cos^3 x sin^2 x +2cos x sin^2 x +sin^2 x Transforming using sin^2 x =1−cos^2 x 16cos^5 x −20cos^3 x −2cos^2 x +5cos x +1=0 16cos^5 x −20cos^3 x +5cos x=2cos^2 x −1 cos 5x =cos 2x But x=((2π)/7) cos ((10π)/7) =cos ((4π)/7) cos α =cos (2π−α) cos ((10π)/7) =cos (2π−((10π)/7)) =cos ((4π)/7)

$$\frac{{a}}{{b}}=\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\Leftrightarrow\:{c}^{\mathrm{2}} =−\frac{{a}^{\mathrm{3}} }{{b}}+{ab}+{b}^{\mathrm{2}} \\ $$$$\mathrm{We}\:\mathrm{know}\:\frac{{a}}{\mathrm{sin}\:\mathrm{4}\theta}=\frac{{b}}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{{c}}{\mathrm{sin}\:\theta} \\ $$$$\mathrm{and}\:\mathrm{4}\theta+\mathrm{2}\theta+\theta=\pi\:\Rightarrow\:\theta=\frac{\pi}{\mathrm{7}} \\ $$$$\mathrm{Let}\:{x}=\mathrm{2}\theta=\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$$\mathrm{Let}\:{a}=\mathrm{sin}\:\mathrm{2}{x}\:\wedge{b}=\mathrm{sin}\:{x}\:\wedge{c}=\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:\Leftrightarrow \\ $$$${a}=\mathrm{2cos}\:{x}\:\mathrm{sin}\:{x}\:\wedge{b}=\mathrm{sin}\:{x}\:\wedge{c}^{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{2}} \\ $$$$\mathrm{Inserting}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{2}}=−\mathrm{8cos}^{\mathrm{3}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:+\mathrm{2cos}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{Transforming}\:\mathrm{using}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{16cos}^{\mathrm{5}} \:{x}\:−\mathrm{20cos}^{\mathrm{3}} \:{x}\:−\mathrm{2cos}^{\mathrm{2}} \:{x}\:+\mathrm{5cos}\:{x}\:+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{16cos}^{\mathrm{5}} \:{x}\:−\mathrm{20cos}^{\mathrm{3}} \:{x}\:+\mathrm{5cos}\:{x}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:−\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{5}{x}\:=\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\mathrm{But}\:{x}=\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{10}\pi}{\mathrm{7}}\:=\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\alpha\:=\mathrm{cos}\:\left(\mathrm{2}\pi−\alpha\right) \\ $$$$\mathrm{cos}\:\frac{\mathrm{10}\pi}{\mathrm{7}}\:=\mathrm{cos}\:\left(\mathrm{2}\pi−\frac{\mathrm{10}\pi}{\mathrm{7}}\right)\:=\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}} \\ $$

Commented by Mingma last updated on 09/Nov/23

Nice!

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