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 February 24th, 2016, 11:22 PM #1 Newbie   Joined: Feb 2016 From: singapore Posts: 3 Thanks: 0 Trig identities Hey, I found an interesting trig identity: (sin18°) (cos36°) = 1/4 I tried using the double angle formula and compound angle, but I could not solve it. Last edited by skipjack; March 23rd, 2016 at 10:32 PM.
 February 25th, 2016, 02:20 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics $\displaystyle \sin 9^\circ=\frac{1}{8} [\sqrt{10}+\sqrt2-2\sqrt{5-\sqrt5}]$ $\displaystyle \cos9^\circ=\frac{1}{8} [\sqrt{10}+\sqrt2+2\sqrt{5-\sqrt5}]$ $\displaystyle \cos 36^\circ=\frac{\sqrt5+1}{4}$ (Source) Now you can derive it Last edited by 123qwerty; February 25th, 2016 at 02:27 AM.
 February 25th, 2016, 02:31 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,285 Thanks: 1681 Using the double angle identity, I proved sin(18°)cos(36°) = 1/4 (and several related results) here. Thanks from 123qwerty
 March 1st, 2016, 02:54 PM #4 Newbie   Joined: Feb 2016 From: singapore Posts: 3 Thanks: 0 Thanks
 March 1st, 2016, 05:48 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond $\displaystyle \cos72^\circ=2\cos^236^\circ-1$ $\displaystyle \sin18^\circ=2\cos^236^\circ-1$ $\displaystyle \sqrt{\dfrac{1-\cos36^\circ}{2}}=2\cos^236^\circ-1$ $\displaystyle \dfrac{1-\cos36^\circ}{2}=4\cos^436^\circ-4\cos^236^\circ+1$ $\displaystyle 8\cos^436^\circ-8\cos^236^\circ+\cos36^\circ+1=0$ $\displaystyle 8x^4-8x^2+x+1=0$ $\displaystyle 8x^2(x^2-1)+x+1=0$ $\displaystyle 8x^2(x-1)(x+1)+x+1=0$ $\displaystyle (x+1)(8x^3-8x^2+1)=0$ $\displaystyle (x+1)(2x-1)(4x^2-2x-1)=0$ $\displaystyle 4x^2-2x-1=0\Rightarrow x=\cos36^\circ=\dfrac{\sqrt5+1}{4}$ $\displaystyle \sin18^\circ\cos36^\circ=\cos72^\circ\cos36^\circ= (2\cos^236^\circ-1)\cos36^\circ=2\cos^336^\circ-\cos36^\circ$ $\displaystyle =2\left(\dfrac{\sqrt5+1}{4}\right)^3-\dfrac{\sqrt5+1}{4}$ $\displaystyle =2\left(\dfrac{5\sqrt5+15+3\sqrt5+1}{64}\right)-\dfrac{\sqrt5+1}{4}$ $\displaystyle =\dfrac{\sqrt5+2}{4}-\dfrac{\sqrt5+1}{4}=\dfrac14$
 March 23rd, 2016, 06:23 AM #6 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry \displaystyle \begin{align*}\sin{18}^\circ\cos{36}^\circ &= \frac{\sin{36}^\circ}{2\cos{18}^\circ} \cos{36}^\circ \\ &= \frac{\frac{1}{2}\sin{72}^\circ}{2\cos{18}^\circ} \\ &= \frac{\sin{72}^\circ}{4\cos{18}^\circ} \\ &= \frac{\sin{72}^\circ}{4\cos{(90-72)}^\circ} \\ &= \frac{\cancel{\sin{72}^\circ}} {4\cancel{\sin{72}^\circ}} \\ &= \frac{1}{4}\end{align*} Thanks from greg1313 Last edited by skipjack; March 23rd, 2016 at 10:30 PM.

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