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Question Number 68673 by mr W last updated on 14/Sep/19

$$find\mathbf{sin}20°=?$$

Commented by MJS last updated on 15/Sep/19

$$\Rightarrow \mathrm{we}\mathrm{cannot}\mathrm{use}\mathrm{this}\mathrm{to}\mathrm{get}\sin 1°,\mathrm{because}$$$$\sin 3°=\frac{1}{8}(1-\sqrt{3})\sqrt{5+\sqrt{5}}-\frac{1}{16}(1-\sqrt{5})(1+\sqrt{3})\sqrt{2}$$$$\mathrm{and}$$$$\sin 3°=3\sin 1°-{4\sin }^{3}1°$$$$\mathrm{also}\mathrm{leads}\mathrm{to}\mathrm{a}\mathrm{super}\mathrm{complex}\mathrm{Cardano}$$$$‘‘{\mathrm{solution}}^{″}\mathrm{which}\mathrm{is}\mathrm{totally}\mathrm{useless}$$$$\mathrm{and}\mathrm{the}\mathrm{trigonometric}\mathrm{method}\mathrm{gives}$$$$x_0=\sin \left(\frac{1}{3}\arcsin \left(\sin 3°\right)\right)=\sin 1°$$$$x_1=\cos (\frac{\pi }{6}+\frac{1}{3}\arcsin \left(\sin 3°\right))=\cos 31°$$$$x_2=-\sin (\frac{\pi }{3}+\frac{1}{3}\arcsin \left(\sin 3°\right))=-\cos 29°$$

Commented by Prithwish sen last updated on 15/Sep/19

$$\mathrm{Thanks}\mathrm{Sir}.$$

Commented by MJS last updated on 14/Sep/19

$$wehaveanexactvaluefor\sin 3°=\sin \frac{\pi }{60}\Rightarrow$$$${we}^{\prime }reabletocalculate\sin \frac{n\pi }{60}\mathrm{\forall }n\in \mathbb{R}butwe$$$$donothave\sin 2°nor\sin 1°\Rightarrow \mathrm{we}\mathrm{cannot}$$$$\mathrm{calculate}\mathrm{any}\mathrm{other}\mathrm{exact}\mathrm{value}...$$

Commented by Maclaurin Stickker last updated on 15/Sep/19

$$Wecancalculatesin\left(1°\right),butittakes$$$$timeandtheresultisverycomplex.$$

Commented by mr W last updated on 15/Sep/19

$$thankyousir!$$$$cardanocanbeappliedonlyif\mathrm{\Delta }⩾0.$$$$butinthiscase\mathrm{\Delta }={(-\frac{1}{4})}^3+{\left(\frac{\sqrt{3}}{16}\right)}^2=-\frac{1}{256}<0$$

Commented by Prithwish sen last updated on 15/Sep/19

$${\mathrm{Sin}60}^{°}={3\mathrm{Sin}20}^{°}-{4\mathrm{Sin}}^3{20}^{°}$$$$\mathrm{Let}{\mathrm{Sin}20}^{°}=\mathrm{x}$$$$\therefore \frac{\sqrt{3}}{2}=3\mathrm{x}-{4\mathrm{x}}^3$$$${\mathrm{x}}^3-\frac{3}{4}\mathrm{x}+\frac{\sqrt{3}}{8}=0$$$$\mathbf{By}{\mathbf{Cardano}}^{\prime }\mathbf{s}\mathbf{method}\mathbf{for}$$$${\mathbf{x}}^3-3\mathbf{px}-2\mathbf{q}=0$$$$\mathbf{x}{=}^3\sqrt{\mathbf{q}+\sqrt{{\mathbf{q}}^2-{\mathbf{p}}^3}}{+}^3\sqrt{\mathbf{q}-\sqrt{{\mathbf{q}}^2-{\mathbf{p}}^3}}$$$$\mathbf{we}\mathbf{get}\mathbf{p}=\frac{1}{4}\mathbf{and}\mathbf{q}=-\frac{\sqrt{3}}{16}$$$$\mathbf{x}={}^3\sqrt{\frac{-\sqrt{3}}{16}+\sqrt{\frac{3}{256}-\frac{1}{64}}}{+}^3\sqrt{\frac{-\sqrt{3}}{16}+\sqrt{\frac{3}{256}-\frac{1}{64}}}$$$$={}^3\sqrt{\frac{-\sqrt{3+}\mathbf{i}}{16}}{+}^3\sqrt{\frac{-\sqrt{3}-\mathbf{i}}{16}}$$$$\mathbf{Sir},\mathbf{waiting}\mathbf{for}\mathbf{your}\mathbf{valuable}\mathbf{comment}.$$

Commented by Prithwish sen last updated on 15/Sep/19

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}!$$

Commented by MJS last updated on 15/Sep/19

$$\mathrm{using}\mathrm{Cardano}\mathrm{is}\mathrm{possible}\mathrm{but}\left(1\right)\mathrm{we}\mathrm{have}\mathrm{to}$$$$\mathrm{approximate}\mathrm{all}3\mathrm{solutions}\mathrm{to}\mathrm{find}\mathrm{which}\mathrm{is}$$$$\mathrm{the}\mathrm{right}\mathrm{one}\mathrm{and}\left(2\right)\mathrm{the}\mathrm{written}\mathrm{out}\mathrm{solution}$$$$\mathrm{is}\mathrm{not}\mathrm{useable}.\mathrm{but}\mathrm{also}\mathrm{the}\mathrm{trigonometric}$$$$\mathrm{method}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{satisfy}\mathrm{sometimes}$$$$(x-2)(x+3)(x-5)=0$$$$x^3-4x^2-11x+30=0$$$$x=t+\frac{4}{3}$$$$t^3-\frac{49}{3}t+\frac{286}{27}=0$$$$\mathrm{Cardano}\mathrm{leads}\mathrm{to}$$$$u=\sqrt[3]{-\frac{143}{27}+\frac{20\sqrt{3}\mathrm{i}}{3}};v=\sqrt[3]{-\frac{143}{27}-\frac{20\sqrt{3}\mathrm{i}}{3}}$$$$x_1=u+v$$$$x_2=(-\frac{1}{2}-\frac{\sqrt{3}\mathrm{i}}{2})u+(-\frac{1}{2}+\frac{\sqrt{3}\mathrm{i}}{2})v$$$$x_3=(-\frac{1}{2}+\frac{\sqrt{3}\mathrm{i}}{2})u+(-\frac{1}{2}-\frac{\sqrt{3}\mathrm{i}}{2})v$$$$\mathrm{and}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{handle}\mathrm{these}$$$$\mathrm{or}\mathrm{to}\mathrm{analytically}\mathrm{show}{x}_1=5;x_2=2;x_3=-3$$$$$$$$\mathrm{the}\mathrm{trigonometric}\mathrm{method}\mathrm{gives}$$$$x_0=\frac{4}{3}+\frac{14}{3}\sin \left(\frac{1}{3}\arcsin \frac{143}{343}\right)$$$$x_1=\frac{4}{3}+\frac{14}{3}\cos (\frac{\pi }{6}+\frac{1}{3}\arcsin \frac{143}{343})$$$$x_2=\frac{4}{3}-\frac{14}{3}\sin (\frac{\pi }{3}+\frac{1}{3}\arcsin \frac{143}{343})$$$$\mathrm{again},\mathrm{try}\mathrm{to}\mathrm{show}{x}_0=2;x_1=5;x_2=-3$$$$$$$$\mathrm{as}\mathrm{we}\mathrm{see},\mathrm{the}\mathrm{actual}\mathrm{solutions}\mathrm{can}\mathrm{be}\mathrm{well}\mathrm{hidden}$$

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