How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?.

Solution 1:

The complex roots of $x^5-1$ are: $$ \begin{align} x_1&=1\\ x_2&=\cos\frac{2\pi}5+i\sin\frac{2\pi}5\\ x_3&=-\cos\frac{\pi}5+i\sin\frac{\pi}5\\ x_4&=-\cos\frac{\pi}5-i\sin\frac{\pi}5\\ x_5&=\cos\frac{2\pi}5-i\sin\frac{2\pi}5 \end{align} $$ using Vieta's formulas you get $$0=x_1+x_2+\dots+x_5=1+2\left(\cos\frac{2\pi}5-\cos\frac\pi5\right)=0,$$ which yields your first equation.

From now on, let $\varphi=\frac{\pi}5$ (for brevity).

We know that $\cos2\varphi-\cos\varphi+\frac12=(2\cos^2\varphi-1)-\cos\varphi+\frac12=2\cos^2\varphi-\cos\varphi-\frac12=0$, i.e. $$2\cos^2\varphi-\cos\varphi=\frac12.$$ Now from $\cos2\varphi=\cos\varphi-\frac12$ you get $$\cos\varphi\cos2\varphi=\cos^2\varphi-\frac{\cos\varphi}2=\frac{2\cos^2\varphi-\cos\varphi}2=\frac14.$$

(Or, as suggested in Chandrasekhar's answer, from $2\cos^2\varphi-\cos\varphi=\frac12$ you can find the value of $\cos\varphi$ by solving the quadratic equation and taking the positive root. Once you know $\cos\varphi$, you can compute $\cos2\varphi$ and many other things. If you try it this way, you can check your result e.g. here.)

Solution 2:

Note that: $\cos{2x} = \cos^{2}{x} - \sin^{2}{x} = 2\:\cos^{2}{x} - 1$. Therefore you have $\cos \frac{2\pi}{5} = 2\:\cos^{2}\frac{\pi}{5} - 1$

Now, \begin{align*} \cos\frac{\pi}{5} - \cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - 2\: \cos^{2}\frac{\pi}{5}+1 \end{align*} This is a quadratic equation of the form $2 x^{2} - x -1 =0$ and solving this will give you the value of $\cos\frac{\pi}{5}$ from which you can find the above value which you need.

Solution 3:

For the first equality.
\begin{split} \cos(\pi/5) - \cos(2\pi/5) &=\cos(3\pi/10-\pi/10) - \cos(3\pi/10+\pi/10)\\ &=2\sin(\pi/10)\sin(3\pi/10)=2\sin(\pi/10)\cos(\pi/5)\\ &=\frac{2\sin(\pi/10)\cos(\pi/10)\cos(\pi/5)}{\cos(\pi/10)}\\ &=\frac{\sin(\pi/5)\cos(\pi/5)}{\cos(\pi/10)}\\ &=\frac{\sin(2\pi/5)}{2\cos(\pi/10)}=\frac 12. \end{split}

Here 's the latter equality: \begin{split} \cos(\pi/5)\cos(2\pi/5) &= \frac{\sin(\pi/5)\cos(\pi/5)\cos(2\pi/5)}{\sin(\pi/5)}\\ &=\frac{\sin(2\pi/5)\cos(2\pi/5)}{2\sin(\pi/5)}\\ &=\frac{ \sin(4\pi/5)}{4\sin(\pi/5)}\\ &=\frac 14.\end{split}

Solution 4:

I was not able (yet) to follow Chandrasekar's solution, but noticed this while trying to understand the argument (how it could possibly lead to the solution, or how exactly he arrives at $2x^2-x-1$ for $x=\cos\frac{\pi}{5}$, which to me seems non-obvious and even a fallacious deduction from his equations and prose -- apologies if I am just being dense)...perhaps it is what Chandrasekar meant all along, but in any case, it does seem to be the most elementary solution available.

Apply the double angle formula $\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1$ to $\theta=\frac{\pi}{5}$ and $\frac{2\pi}{5}$, with $a=\cos\frac{\pi}{5}$ and $b=\frac{2\pi}{5}$ for convenience, recalling also that $\cos(\pi\pm\theta)=-\cos\theta$: $$ b=\cos\frac{2\pi}{5}=2\,\cos^2\frac{\pi}{5}-1=2a^2-1 $$ $$ -a=\cos\frac{4\pi}{5}=2\,\cos^2\frac{2\pi}{5}-1=2b^2-1 $$ Next, subtracting the equations $$ \matrix{ 2a^2=1+b\\ 2b^2=1-a} $$ we get

$$ \eqalign{ 2\left(a^2-b^2\right)&=b+a\\ 2\left(a+b\right)\left(a-b\right)&=b+a\\ 2\left(a-b\right)&=1\\ a-b&=\frac12\,. } $$ Furthermore, multiplying, we get $$ 4(ab)^2=(1+b)(1-a)=1+(b-a)-ab=1+\left(-\tfrac12\right)-ab $$ giving us the quadratic equation $$4(ab)^2+(ab)-\tfrac12=0$$ $$8(ab)^2+2(ab)-1=0$$ $$\left(4ab-1\right)\left(2ab+1\right)=0$$ so that $ab=\frac14$ or $-\frac12$, from which we can choose the former since we know that $0&lta&ltb$.

Solution 5:

$$\cos(\pi/5)\cos(2\pi/5)=A$$

$$\Longrightarrow\quad A = \frac{\sin(\pi/5)\cos(\pi/5)\cos(2\pi/5)}{\sin(\pi/5)}=\frac{\sin(2\pi/5)\cos(2\pi/5)}{2\sin(\pi/5)}$$

$$A = \frac{\sin(4\pi/5)} {2\cdot 2\cdot\sin(\pi/5)}=\frac{1}{4}$$ since $\sin(4\pi/5)=\sin(\pi/5)$.