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Given $4 \cos^2 x -2\cos x -1 = 0$.

Use this to show that $\cos 36^{\circ} = (1+ \sqrt 5)/4$, $\cos 72^{\circ} = (-1+\sqrt 5)/4$

Your help is greatly appreciated! Thanks

Ron Gordon
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Teddy
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  • Hmm... so it appears that $(2\cos x-1)^2=\tfrac 54$ from the first statement. What have you tried so far? – abiessu Sep 19 '13 at 13:18

3 Answers3

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To derive this from fundamentals, note that

$$\sin{108^{\circ}} = \sin{72^{\circ}}$$

then use a double-angle and triple-angle forumla:

$$\sin{2 x} = 2 \sin{x} \cos{x}$$ $$\sin{3 x} = 3 \sin{x} - 4 \sin^3{x}$$

In this case, $x=36^{\circ}$. Setting the above two equations equal to each other results in the quadratic equation in question:

$$2 \cos{x} = 3 - 4 (1-\cos^2{x}) = 4 \cos^2{x}-1$$

The rest follows from the above discussion.

Ron Gordon
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  • I believe that only this answer is right. To make the first answer valid, you need to make also reasoning that cos is a function on $[0;Pi]$. The solution suggested by Ron is quite universal, as I think it should work for any k, not only for 5 (36 = $Pi/5$). – outmind Apr 22 '17 at 21:04
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Hint: Look at the Quadratic Formula:

The solution to $ax^2+bx+c=0$ is $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

The equation is based on the fact that $$ \cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x) $$ and that $\cos(5\cdot36^\circ)=-1$ to get $$ 16\cos^5(36^\circ)-20\cos^3(36^\circ)+5\cos(36^\circ)+1=0 $$ Factoring yields $$ (\cos(36^\circ)+1)(4\cos^2(36^\circ)-2\cos(36^\circ)-1)^2=0 $$ We know that $\cos(36^\circ)+1\ne0$; therefore, $$ 4\cos^2(36^\circ)-2\cos(36^\circ)-1=0 $$ Deciding between the two roots of this equation is a matter of looking at the signs of the roots.

For $\cos(72^\circ)$, use the identity $\cos(2x)=2\cos^2(x)-1$.

robjohn
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    Applying quad formula I do get (1+5√)/4, but how does the 36 degrees come in? – Teddy Sep 19 '13 at 13:30
  • Use the inverse cosine function arccos.... $x=\arccos{\frac{1+\sqrt{5}}{4}}$..... – Eleven-Eleven Sep 19 '13 at 13:37
  • OH, I got it alr. Thanks. – Teddy Sep 19 '13 at 13:39
  • @ChristopherErnst That looks like a circular argument to me. (Why is $36^\circ = \arccos\frac{1 + \sqrt{5}}{4}$?) Also, if you had access to $\arccos$ all along, why bother with the quadratic equation? – Arthur Sep 19 '13 at 13:42
  • @Teddy: I hope the last addition to my answer makes things look less circular (although $\cos$ is a circular function). – robjohn Sep 19 '13 at 16:13
  • how is it circular? you substitute $y=\cos{x}$ You solve the quadratic and get $\cos{x}=\frac{1+\sqrt{5}}{4}$. Then you solve for $x$...how is that circular? – Eleven-Eleven Sep 19 '13 at 16:18
  • @ChristopherErnst: First of all, it was a play on words since Teddy mentioned a circular argument and $\cos$ is a circular function. However, I think he was wondering why $36^\circ$ satisfied $4\cos^2(x)-2\cos(x)-1=0$ since that was never explained. – robjohn Sep 19 '13 at 16:25
  • So I missed the joke.... Darn.... I try not to be that guy... :) – Eleven-Eleven Sep 19 '13 at 17:06
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Using $\cos5\theta = 16\cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$;

$\cos 180 = 16\cos^5 36 - 20 \cos^3 36 + 5 \cos 36$

Let $\cos36 = x$:

$-1 = 16x^5 -20x^3 +5x$

Its solutions are ${-1,\frac{1- \sqrt{5}}{4}},\frac{1+ \sqrt{5}}{4}$

And as $\cos36 = x$, $\cos 36$ must be equal to one of them.

$\cos 36$ must be equal to $\frac{1+\sqrt{5}}{4}$, as $-1$ and $\frac{1- \sqrt{5}}{4}$ are negative.

MCCCS
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