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 February 29th, 2012, 12:36 AM #1 Newbie   Joined: Feb 2012 Posts: 17 Thanks: 0 De Moivre's Theorem how to find the exact value sin??2?/5? using Demoivers Theorem algebraically ?
 February 29th, 2012, 01:31 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: De Moivre's Theorem $$$\cos\(\frac{2\pi}{5}$$+i\cdot\sin$$\frac{2\pi}{5 }$$\)^5=\cos$$2\pi$$+i\cdot\sin$$2\pi$$=1$ We may then consider the roots of the equation: $x^5-1=0$ $\frac{1}{4}(x-1)$$2x^2+\(1+\sqrt{5}$$x+2\)$$2x^2+\(1-\sqrt{5}$$x+2\)=0$ Now simply choose the complex root a + bi where a and b are positive, i.e., in the first quadrant of the complex plane. This will come from the quadratic factor whose coefficient of the linear term is negative: $2x^2+$$1-\sqrt{5}$$x+2=0$ and choosing the root $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$ we find: $x=\frac{\sqrt{5}-1+\sqrt{$$1-\sqrt{5}$$^2-4(2)(2)}}{2(2)}=\frac{\sqrt{5}-1}{4}+i\frac{\sqrt{2$$5+\sqrt{5}$$}}{4}$ Thus: $\cos$$\frac{2\pi}{5}$$=\frac{\sqrt{5}-1}{4}$ $\sin$$\frac{2\pi}{5}$$=\frac{\sqrt{2$$5+\sqrt{5}$$ }}{4}$

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