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The problem reads as follows:

Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of $$\cos(36^\circ)$$

it says that you may find useful the following identities: $$\cos^2 t+\sin^2 t = 1,\\ \sin 2t = 2\sin t\cos t,\\ \sin 3t = 3\sin t - 4\sin^3 t. $$

Do I have to do a system of linear equations in function of ..what? $t$? $\cos$?

Thanks in advance :)

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In the third century BC, Euclid essentially found the sine of $36^\circ$. See my answer below. – Michael Hardy Jun 13 '14 at 13:30

Let $t=\frac{\pi}{5}$ (so $t$ is $36^\circ$). Since $108=180-72$, we have $3t=\pi-2t$ and therefore $$\sin(3t)=\sin(\pi-2t).$$ But $\sin(\pi-2t)=\sin(2t)=2\sin t\cos t$.

Also, by the identity you were given, $\sin(3t)=3\sin t-4\sin^3 t$. Thus $$3\sin t-4\sin^3 t=2\sin t\cos t.$$ But $\sin t\ne 0$, so we can cancel a $\sin t$ and obtain $$3-4\sin^2 t=2\cos t.$$ Substitute $1-\cos^2 t$ for $\sin^2 t$ and simplify a bit. We get $$4\cos^2 t-2\cos t-1=0.$$ Use the Quadratic Formula to solve this quadratic equation for $\cos t$, rejecting the negative root. We get $$\cos t=\frac{2+\sqrt{20}}{8}.$$ We can simplify this to $\dfrac{1+\sqrt{5}}{4}.$

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Very well explained, thanks very much :) deeply appreciated – Maximilian1988 Feb 11 '13 at 8:31

To find $\cos{\pi/5}$, note that

$$\sin{(3 \pi/5)} = \sin{(2 \pi/5)}$$


$$\sin{(3 \pi/5)} = 3 \sin{(\pi/5)} - 4 \sin^3{(\pi/5)} = 2 \sin{(\pi/5)} \cos{(\pi/5)}$$


$$2 \cos{(\pi/5)} = 3 - 4 \sin^2{(\pi/5)} = 4 \cos^2{(\pi/5)} - 1$$

Let $y=\cos{(\pi/5)}$. Then

$$4 y^2-2 y-1=0 \implies y = \frac{1 + \sqrt{5}}{4}$$

because $y>0$. Thus, $\cos{(\pi/5)} = (1+\sqrt{5})/4$.

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Thanks very much Rlgordonma :) – Maximilian1988 Feb 11 '13 at 8:31
This is definitely the route the hints were indicating (+1). – robjohn Feb 11 '13 at 9:02

Since rlgordonma has given the answer suggested by the supplied hint, here is another method of computing $\cos(\pi/5)$.

In this answer, it is shown that $$ \tan(5\theta)=\frac{5\tan(\theta)-10\tan^3(\theta)+\tan^5(\theta)}{1-10\tan^2(\theta)+5\tan^4(\theta)} $$ which, since $\tan(\pi/2)=\infty$, implies that $$ 5\tan^4(\pi/10)-10\tan^2(\pi/10)+1=0 $$ which, by the quadratic formula, yields $$ \tan^2(\pi/10)=\frac{5-2\sqrt{5}}{5} $$ Adding $1$ and taking the reciprocal yields $$ \frac{1+\cos(\pi/5)}{2}=\cos^2(\pi/10)=\frac{5+\sqrt5}{8} $$ Therefore, $$ \cos(\pi/5)=\frac{1+\sqrt5}{4}=\phi/2 $$ where $\phi$ is the Golden Ratio.

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I love solutions invoking the golden ratio. It'd be even better with a pentagram. (+1) – Ron Gordon Feb 11 '13 at 9:05

Here is a way using roots of unity.

We have $\cos 36 = \frac{\omega + \omega^{-1}}{2}$ where $\omega = \text{exp} \left ( \frac{2 \pi}{10} \right )$. We have $-w$ is a primitive $5^{th}$ root of unity, so it follows $\omega^4 - \omega^3 + \omega^2 - \omega + 1 = 0$. Now, $x^4 - x^3 + x^2 - x + 1 = x^2(x^2 - x + 1 - \frac{1}{x} + \frac{1}{x^2}) = x^2\left ( \left (x + \frac{1}{x} \right )^2 - \left (x + \frac{1}{x} \right ) -1 \right )$. Thus $0 = \omega^2 \cdot ((2 \cos 36)^2 - (2 \cos 36) - 1)$. Therefore $4 \cos^3 36 - 2 \cos 36 - 1 = 0$. Using the quadratic formula we then arrive at $$\cos 36 = \frac{1 + \sqrt{5}}{4}$$

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I have never ever done roots of unity.. thanks though for your time and help – Maximilian1988 Feb 11 '13 at 8:23
@dinoboy: I think you made a mistake somewhere. – Ron Gordon Feb 11 '13 at 8:24
Can you elaborate on where? Considering we got the same answer I'm not sure what drove you to make that claim. – dinoboy Feb 11 '13 at 8:24
@dinoboy, we do? Oh, I see, you wrote yours down wrong. – Ron Gordon Feb 11 '13 at 8:30
Oh yes indeed, I made a typo. – dinoboy Feb 11 '13 at 15:40

This is a graphic that I created for my website about 2 days ago. Using $\Delta ABC$ invoke the Law of Cosines ($\angle BAC = \angle BCA = 36^\circ$).

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BEGIN QUOTE: Ptolemy used geometric reasoning based on Proposition 10 of Book XIII of Euclid's Elements to find the chords of $72^\circ$ and $36^\circ$. That Proposition states that if an equilateral pentagon is inscribed in a circle, then the area of the square on the side of the pentagon equals the sum of the areas of the squares on the sides of the hexagon and the decagon inscribed in the same circle. END QUOTE

Finding the chord of $72^\circ$ amounts to finding the sine of $36^\circ$.

Here is Proposition  10 of Book VIII.

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Another with roots of unity ...

Let $\omega:=\exp\left(\frac{\pi i}{5}\right)$ and $\phi$ is the golden ratio defined via relation $$\phi^2:=\phi+1\qquad|\quad\phi>0\tag{a}$$i.e., explicitly : $\phi=\frac{1+\sqrt{5}}{2}$ (but only (a) is sufficient to proceed further through this text, without radicals of 5's at start), then following holds true :

Via Euler's formula: $$\Re\{\omega\}=\Re\bigg\{ \exp\left(\frac{\pi i}{5}\right) \bigg\} = \cos \left(\frac{\pi}{5}\right)$$


$$\Im\{\omega\}=\sin \left(\frac{\pi}{5}\right)$$

Moreover, it's clear, by properities of the function cosine (resp. sine) on the interval $(0,\frac{\pi}{2})$, that $$\Re\{\omega\}>0\qquad\mathrm{resp.}\qquad\Im\{\omega\}>0\tag{0}$$ By Moivre's formula also $$\omega^5 = e^{i\pi}=-1$$ I.e.: $$\omega^5+1=0$$

Since $\omega\neq -1$ we divide this by factor $\omega+1$, hence


This could be wieved as a polynomial degree 4 in $\omega$, so we may guess, it can be factorised more suggesting :

$$\omega^4-\omega^3+\omega^2-\omega+1 = (\omega^2-\alpha\omega+1)(\omega^2-\beta\omega+1)\tag{2}$$ Expanding out we get a system for $\alpha$ and $\beta$ :

$$\alpha+\beta=1$$ $$\alpha\beta=-1$$ Subctracting $\phi$ in the equation for golden ratio and dividing through $\phi$ we get an equivalent relation $$\phi-\frac{1}{\phi}=1\tag{b}$$

So the conditions for $\alpha$ and $\beta$ are met when :

$$\alpha=\phi$$ $$\beta=-\frac{1}{\phi}$$

Now we have from $(1)$ and $(2)$ two quadratic equations for $\omega$ :

$$\omega=\frac{\alpha}{2}\pm\frac{1}{2}\sqrt{\alpha^2-4}\qquad \mathrm{or}\qquad \omega=\frac{\beta}{2}\pm\frac{1}{2}\sqrt{\beta^2-4}\tag{3}$$

Dividing $(b)$ by $\phi$ and rearanging we get $$\frac{1}{\phi}=1-\frac{1}{\phi^2}\tag{c}$$

So then $$\alpha^2-4=\phi^2-4\overset{(a)}{=}\phi-3\overset{(b)}{=}\frac{1}{\phi}-2\overset{(c)}{=}-\frac{1}{\phi^2}-1<0 \quad\mathrm{since}\quad \phi>0$$

... and also ... $$\beta^2-4=\frac{1}{\phi^2}-4\overset{(c)}{=}-\frac{1}{\phi}-3<0 \quad\mathrm{since}\quad \phi>0$$

Taking square roots make us some imaginary numbers, from (3) then, taking real part

$$\Re\{\omega\}=\frac{\alpha}{2}\qquad \mathrm{or}\qquad \Re\{\omega\}=\frac{\beta}{2}$$

However, since $\beta=-\frac{1}{\phi}<0$ is by $(0)$ and $(3)$ nessesarily and uniquely :




and, as a bonus :


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