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

Answered by deleteduser1 last updated on 09/Nov/23

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

Commented by Mingma last updated on 09/Nov/23

Nice, sir!

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

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

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

$$Butsirsince\sin \left(\frac{\pi }{7}\right)hasno‘‘{nice}^{″}$$$$value,so{there}^{\prime }sno‘‘nice{solution}^{″}$$$$ofthequestion?$$

Commented by Frix last updated on 09/Nov/23

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

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

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathrm{dear}\mathbf{sir}\mathrm{for}\mathrm{useful}\mathrm{hint}.$$

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

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

Commented by Mingma last updated on 09/Nov/23

Nice!

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