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Question Number 121256 by ZiYangLee last updated on 06/Nov/20

$$\mathrm{Prove}{2\sin }^{2}3\theta -{2\sin }^2\theta =\cos 2\theta -\cos 6\theta .$$$$\mathrm{By}\mathrm{substitusing}\theta =\frac{\pi }{10}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{identity},$$$$\mathrm{prove}\mathrm{that}\sin \frac{3\pi }{10}-\sin \frac{\pi }{10}=\frac{1}{2}$$

Answered by TANMAY PANACEA last updated on 06/Nov/20

$$2{sin}^{2}3\theta -2{sin}^2\theta$$$$1-cos6\theta -1+cos2\theta$$$$cos2\theta -cos6\theta$$$$sosin3\theta -sin\theta =\frac{cos2\theta -cos6\theta }{2(sin3\theta +sin\theta })$$$$sin\frac{3\pi }{10}-sin\frac{\pi }{10}$$$$\frac{cos\frac{2\pi }{10}-cos\frac{6\pi }{10}}{2(sin\frac{3\pi }{10}+sin\frac{\pi }{10})}$$$$\frac{1}{2}\times \frac{2sin\left(\frac{4\pi }{10}\right)sin\left(\frac{2\pi }{20}\right)}{2sin\left(\frac{2\pi }{10}\right)cos\left(\frac{\pi }{10}\right)}=\frac{1}{2}proved$$$$note$$$$sin\frac{4\pi }{10}=cos(\frac{\pi }{2}-\frac{4\pi }{10})=cos\left(\frac{\pi }{10}\right)$$

Commented by ZiYangLee last updated on 08/Nov/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$$$$$

Commented by TANMAY PANACEA last updated on 08/Nov/20

$$mostwelcome$$

Answered by Dwaipayan Shikari last updated on 06/Nov/20

$$sin\frac{3\pi }{10}-\mathit{s}\mathit{i}\mathit{n}\frac{\mathit{\pi }}{10}$$$$=\frac{\sqrt{5}+1}{4}-\frac{\sqrt{5}-1}{4}=\frac{1}{2}$$

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