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 
$\displaystyle \sin 9^\circ=\frac{1}{8} [\sqrt{10}+\sqrt22\sqrt{5\sqrt5}]$ $\displaystyle \cos9^\circ=\frac{1}{8} [\sqrt{10}+\sqrt2+2\sqrt{5\sqrt5}]$ $\displaystyle \cos 36^\circ=\frac{\sqrt5+1}{4}$ (Source) Now you can derive it Last edited by 123qwerty; February 25th, 2016 at 02:27 AM. 
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: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond 
$\displaystyle \cos72^\circ=2\cos^236^\circ1$ $\displaystyle \sin18^\circ=2\cos^236^\circ1$ $\displaystyle \sqrt{\dfrac{1\cos36^\circ}{2}}=2\cos^236^\circ1$ $\displaystyle \dfrac{1\cos36^\circ}{2}=4\cos^436^\circ4\cos^236^\circ+1$ $\displaystyle 8\cos^436^\circ8\cos^236^\circ+\cos36^\circ+1=0$ $\displaystyle 8x^48x^2+x+1=0$ $\displaystyle 8x^2(x^21)+x+1=0$ $\displaystyle 8x^2(x1)(x+1)+x+1=0$ $\displaystyle (x+1)(8x^38x^2+1)=0$ $\displaystyle (x+1)(2x1)(4x^22x1)=0$ $\displaystyle 4x^22x1=0\Rightarrow x=\cos36^\circ=\dfrac{\sqrt5+1}{4}$ $\displaystyle \sin18^\circ\cos36^\circ=\cos72^\circ\cos36^\circ= (2\cos^236^\circ1)\cos36^\circ=2\cos^336^\circ\cos36^\circ$ $\displaystyle =2\left(\dfrac{\sqrt5+1}{4}\right)^3\dfrac{\sqrt5+1}{4}$ $\displaystyle =2\left(\dfrac{5\sqrt5+15+3\sqrt5+1}{64}\right)\dfrac{\sqrt5+1}{4}$ $\displaystyle =\dfrac{\sqrt5+2}{4}\dfrac{\sqrt5+1}{4}=\dfrac14$ 
March 23rd, 2016, 06:23 AM  #6 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
$\displaystyle \begin{align*}\sin{18}^\circ\cos{36}^\circ &= \frac{\sin{36}^\circ}{2\cos{18}^\circ} \cos{36}^\circ \\ &= \frac{\frac{1}{2}\sin{72}^\circ}{2\cos{18}^\circ} \\ &= \frac{\sin{72}^\circ}{4\cos{18}^\circ} \\ &= \frac{\sin{72}^\circ}{4\cos{(9072)}^\circ} \\ &= \frac{\cancel{\sin{72}^\circ}} {4\cancel{\sin{72}^\circ}} \\ &= \frac{1}{4}\end{align*}$ Last edited by skipjack; March 23rd, 2016 at 10:30 PM. 

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