Question Number 199771 by Rupesh123 last updated on 09/Nov/23
Answered by AST last updated on 09/Nov/23
$$\frac{{sin}\left(\mathrm{4}{x}\right)=\mathrm{2}{sin}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{2}{x}\right)}{{a}}=\frac{{sin}\left(\mathrm{2}{x}\right)}{{b}} \\ $$$$\Rightarrow{cos}\left(\mathrm{2}{x}\right)=\frac{{a}}{\mathrm{2}{b}}=\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\Rightarrow{cos}^{\mathrm{2}} {x}=\frac{{a}+\mathrm{2}{b}}{\mathrm{4}{b}} \\ $$$$\frac{{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)}{{b}}=\frac{{sin}\left({x}\right)}{{c}}\Rightarrow{cos}\left({x}\right)=\frac{{b}}{\mathrm{2}{c}} \\ $$$$\Rightarrow\frac{{a}+\mathrm{2}{b}}{{b}}=\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\Rightarrow\frac{{a}}{{b}}=\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }−\mathrm{2} \\ $$
Commented by Mingma last updated on 09/Nov/23
Nice, sir!
Answered by Rasheed.Sindhi last updated on 09/Nov/23
$$\frac{{a}}{\mathrm{sin}\:\mathrm{4}\theta}=\frac{{b}}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{{c}}{\mathrm{sin}\:\theta}={k}\:\left({say}\right) \\ $$$${a}={k}\:\mathrm{sin}\:\mathrm{4}\theta,{b}={k}\:\mathrm{sin}\:\mathrm{2}\theta,{c}={k}\:\mathrm{sin}\:\theta \\ $$$$\frac{{a}}{{b}}=\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\Rightarrow \\ $$$$\:\:\:\:\frac{{k}\:\mathrm{sin}\:\mathrm{4}\theta}{{k}\:\mathrm{sin}\:\mathrm{2}\theta}=\frac{{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta\:}{{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{4}\theta−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$\:\:\:\:\frac{\mathrm{sin}\:\mathrm{4}\theta}{\:\mathrm{sin}\:\mathrm{2}\theta}=\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{sin}^{\mathrm{2}} \theta\:}{\mathrm{sin}^{\mathrm{2}} \mathrm{4}\theta−\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$${lhs}:\:\:\frac{\mathrm{sin}\:\mathrm{4}\theta}{\:\mathrm{sin}\:\mathrm{2}\theta}=\frac{\mathrm{2}\:\cancel{\mathrm{sin}\:\mathrm{2}\theta}\:\mathrm{cos}\:\mathrm{2}\theta}{\cancel{\mathrm{sin}\:\mathrm{2}\theta}}=\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$${rhs}:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{sin}^{\mathrm{2}} \theta\:}{\mathrm{sin}^{\mathrm{2}} \mathrm{4}\theta−\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$…. \\ $$
Commented by Rasheed.Sindhi last updated on 09/Nov/23
$${But}\:{sir}\:{since}\:\mathrm{sin}\left(\frac{\pi}{\mathrm{7}}\right)\:{has}\:{no}\:“{nice}'' \\ $$$${value},\:{so}\:{there}'{s}\:{no}\:“{nice}\:{solution}'' \\ $$$${of}\:{the}\:{question}? \\ $$
Commented by Frix last updated on 09/Nov/23
$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\left(\mathrm{4}+\mathrm{2}+\mathrm{1}\right)\theta=\pi\:\Rightarrow\:\theta=\frac{\pi}{\mathrm{7}} \\ $$$$\mathrm{Should}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{show}… \\ $$
Commented by Rasheed.Sindhi last updated on 09/Nov/23
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\mathrm{dear}\:\boldsymbol{\mathrm{sir}}\:\mathrm{for}\:\mathrm{useful}\:\mathrm{hint}. \\ $$
Commented by Frix last updated on 09/Nov/23
$$\mathrm{I}\:\mathrm{found}\:\mathrm{a}\:\mathrm{way},\:\mathrm{see}\:\mathrm{my}\:\mathrm{answer}. \\ $$
Commented by Mingma last updated on 09/Nov/23
Nice, sir!
Answered by Frix last updated on 09/Nov/23
$$\frac{{a}}{{b}}=\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\Leftrightarrow\:{c}^{\mathrm{2}} =−\frac{{a}^{\mathrm{3}} }{{b}}+{ab}+{b}^{\mathrm{2}} \\ $$$$\mathrm{We}\:\mathrm{know}\:\frac{{a}}{\mathrm{sin}\:\mathrm{4}\theta}=\frac{{b}}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{{c}}{\mathrm{sin}\:\theta} \\ $$$$\mathrm{and}\:\mathrm{4}\theta+\mathrm{2}\theta+\theta=\pi\:\Rightarrow\:\theta=\frac{\pi}{\mathrm{7}} \\ $$$$\mathrm{Let}\:{x}=\mathrm{2}\theta=\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$$\mathrm{Let}\:{a}=\mathrm{sin}\:\mathrm{2}{x}\:\wedge{b}=\mathrm{sin}\:{x}\:\wedge{c}=\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:\Leftrightarrow \\ $$$${a}=\mathrm{2cos}\:{x}\:\mathrm{sin}\:{x}\:\wedge{b}=\mathrm{sin}\:{x}\:\wedge{c}^{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{2}} \\ $$$$\mathrm{Inserting}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{2}}=−\mathrm{8cos}^{\mathrm{3}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:+\mathrm{2cos}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{Transforming}\:\mathrm{using}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{16cos}^{\mathrm{5}} \:{x}\:−\mathrm{20cos}^{\mathrm{3}} \:{x}\:−\mathrm{2cos}^{\mathrm{2}} \:{x}\:+\mathrm{5cos}\:{x}\:+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{16cos}^{\mathrm{5}} \:{x}\:−\mathrm{20cos}^{\mathrm{3}} \:{x}\:+\mathrm{5cos}\:{x}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:−\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{5}{x}\:=\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\mathrm{But}\:{x}=\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{10}\pi}{\mathrm{7}}\:=\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\alpha\:=\mathrm{cos}\:\left(\mathrm{2}\pi−\alpha\right) \\ $$$$\mathrm{cos}\:\frac{\mathrm{10}\pi}{\mathrm{7}}\:=\mathrm{cos}\:\left(\mathrm{2}\pi−\frac{\mathrm{10}\pi}{\mathrm{7}}\right)\:=\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}} \\ $$
Commented by Mingma last updated on 09/Nov/23
Nice!