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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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oooouuups sorry sorry.... t= pi/5 so t= pi divided by 5 –  Maximilian1988 Feb 11 '13 at 8:18
I fixed that for you. For the future - see the 'edit'-link underneath your post. Click it! Edit: It may be the case that the link is invisible to a new user such as you, but IIRC you should be able to edit your own posts. –  Jyrki Lahtonen Feb 11 '13 at 8:22
@Maximilian1988 You should accept the answer that you found most helpful. You can accept an answer by clicking on the check-mark next to the answer, right below the up/down vote count. I see that you have never accepted any answer, and thought you might not know how to. –  Paresh Feb 11 '13 at 8:49
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4 Answers

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
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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
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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
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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
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