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February 24th, 2016, 11:22 PM   #1
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Cool 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.
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February 25th, 2016, 02:20 AM   #2
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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.
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February 25th, 2016, 02:31 AM   #3
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Using the double angle identity, I proved sin(18°)cos(36°) = 1/4 (and several related results) here.
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March 1st, 2016, 02:54 PM   #4
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March 1st, 2016, 05:48 PM   #5
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$\displaystyle \cos72^\circ=2\cos^236^\circ-1$

$\displaystyle \sin18^\circ=2\cos^236^\circ-1$

$\displaystyle \sqrt{\dfrac{1-\cos36^\circ}{2}}=2\cos^236^\circ-1$

$\displaystyle \dfrac{1-\cos36^\circ}{2}=4\cos^436^\circ-4\cos^236^\circ+1$

$\displaystyle 8\cos^436^\circ-8\cos^236^\circ+\cos36^\circ+1=0$

$\displaystyle 8x^4-8x^2+x+1=0$

$\displaystyle 8x^2(x^2-1)+x+1=0$

$\displaystyle 8x^2(x-1)(x+1)+x+1=0$

$\displaystyle (x+1)(8x^3-8x^2+1)=0$

$\displaystyle (x+1)(2x-1)(4x^2-2x-1)=0$

$\displaystyle 4x^2-2x-1=0\Rightarrow x=\cos36^\circ=\dfrac{\sqrt5+1}{4}$

$\displaystyle \sin18^\circ\cos36^\circ=\cos72^\circ\cos36^\circ= (2\cos^236^\circ-1)\cos36^\circ=2\cos^336^\circ-\cos36^\circ$

$\displaystyle =2\left(\dfrac{\sqrt5+1}{4}\right)^3-\dfrac{\sqrt5+1}{4}$

$\displaystyle =2\left(\dfrac{5\sqrt5+15+3\sqrt5+1}{64}\right)-\dfrac{\sqrt5+1}{4}$

$\displaystyle =\dfrac{\sqrt5+2}{4}-\dfrac{\sqrt5+1}{4}=\dfrac14$
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March 23rd, 2016, 06:23 AM   #6
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Math Focus: Trigonometry
$\displaystyle \begin{align*}\sin{18}^\circ\cos{36}^\circ &= \frac{\sin{36}^\circ}{2\cos{18}^\circ} \cos{36}^\circ \\
&= \frac{\frac{1}{2}\sin{72}^\circ}{2\cos{18}^\circ} \\
&= \frac{\sin{72}^\circ}{4\cos{18}^\circ} \\
&= \frac{\sin{72}^\circ}{4\cos{(90-72)}^\circ} \\
&= \frac{\cancel{\sin{72}^\circ}} {4\cancel{\sin{72}^\circ}} \\
&= \frac{1}{4}\end{align*}$
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Last edited by skipjack; March 23rd, 2016 at 10:30 PM.
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