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Question Number 139724 by mathlove last updated on 30/Apr/21

$$withoutcientopiccalculeter$$$$sin9=?$$

Commented by Dwaipayan Shikari last updated on 30/Apr/21

$$sin\left(9°\right)?$$

Commented by malwan last updated on 30/Apr/21

$$sin9=\frac{1}{8}\sqrt{2}(1+\sqrt{5})-$$$$\frac{1}{8}(\sqrt{5}-1)\sqrt{5+\sqrt{5}}$$$$=\frac{1}{2}\sqrt{2-\sqrt{\frac{1}{2}(5+\sqrt{5})}}$$

Answered by Dwaipayan Shikari last updated on 30/Apr/21

$$sin\left(\frac{\pi }{10}\right)=\frac{\sqrt{5}-1}{4}$$$$2sin\left(\frac{\pi }{20}\right)\sqrt{1-{sin}^2\left(\frac{\pi }{20}\right)}=\frac{\sqrt{5}-1}{4}$$$$\Rightarrow {sin}^2\left(\frac{\pi }{20}\right)-{sin}^4\left(\frac{\pi }{20}\right)={\left(\frac{\sqrt{5}-1}{8}\right)}^2$$$$\Rightarrow {sin}^4\left(\frac{\pi }{20}\right)-{sin}^2\left(\frac{\pi }{20}\right)+\frac{3-\sqrt{5}}{32}=0$$$$\Rightarrow {sin}^2\left(\frac{\pi }{20}\right)=\frac{1\pm \sqrt{1-\frac{3-\sqrt{5}}{8}}}{2}=\frac{1\pm \sqrt{\frac{5+\sqrt{5}}{8}}}{2}$$$$sin\left(\frac{\pi }{20}\right)=\sqrt{\frac{1-\frac{1}{2}\sqrt{\sqrt{5}\varphi }}{2}}=\frac{1}{2}\sqrt{2-{\left(5\right)}^{1/4}\sqrt{\varphi }}\varphi =GoldenRatio$$

Commented by mathlove last updated on 01/May/21

$$whatis\mathrm{\varnothing }?$$

Commented by Dwaipayan Shikari last updated on 01/May/21

$$GoldenRatio\varphi =\frac{1+\sqrt{5}}{2}=1.618..$$

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