How Can One Prove $\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7) = 1/2$ [duplicate]

Hint: start with $e^{i\frac{\pi}{7}} = \cos(\pi/7) + i\sin(\pi/7)$ and the fact that the lhs is a 7th root of -1.

Let $u = e^{i\frac{\pi}{7}}$, then we want to find $\Re(u + u^3 + u^5)$.

Then we have $u^7 = -1$ so $u^6 - u^5 + u^4 - u^3 + u^2 -u + 1 = 0$.

Re-arranging this we get: $u^6 + u^4 + u^2 + 1 = u^5 + u^3 + u$.

If $a = u + u^3 + u^5$ then this becomes $u a + 1 = a$, and rearranging this gives $a(1 - u) = 1$, or $a = \dfrac{1}{1 - u}$.

So all we have to do is find $\Re\left(\dfrac{1}{1 - u}\right)$.

$\dfrac{1}{1 - u} = \dfrac{1}{1 - \cos(\pi/7) - i \sin(\pi/7)} = \dfrac{1 - \cos(\pi/7) + i \sin(\pi/7)}{2 - 2 \cos(\pi/7)}$

so

$\Re\left(\dfrac{1}{1 - u}\right) = \dfrac{1 - \cos(\pi/7)}{2 - 2\cos(\pi/7)} = \dfrac{1}{2} $

You could obtain this by first multiplying $\cos(\pi/7)+\cos(3\pi/7)+\cos(5\pi/7)$ by $2\sin(\pi/7)$ and then applying the double angle formula for $\sin$ and sum to product formula for $\cos x \sin y$.

Let $a=\pi/7$, $b=3\pi/7$, and $c=5\pi/7$.

Using $$ 2\cos x\sin x =\sin 2x $$ and $$ \cos x\sin y= {\sin(x+y)-\sin(x-y)\over2 } $$ we have $$ \eqalign{ (\cos a+\cos b+\cos c)\cdot 2\sin a &=\color{maroon}{2\cos a\sin a}+\color{darkgreen}{2\cos b\sin a} +\color{darkblue}{2\cos c\sin a}\cr &=\color{maroon}{\sin 2a }+\color{darkgreen}{\sin(b+a)-\sin(b-a)}+\color{darkblue}{ \sin(c+a)-\sin(c-a)}\cr &=\color{teal}{\sin(2\pi/7) }+ {\color{purple}{\sin(4\pi/7)}\color{teal}{-\sin(2\pi/7)}}+ { \sin(6\pi/7)-\color{purple}{\sin(4\pi/7)}}\cr &=\sin(6\pi/7)\cr &=\sin (\pi/7)\cr &=\sin a. } $$ Now divide both sides of $$ (\cos a+\cos b+\cos c)\cdot 2\sin a =\sin a $$ by $2\sin a$.

To elaborate on Mathlover's comment, the three numbers $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$, and $\cos\frac{5\pi}{7}$ are the three roots of the monic Chebyshev polynomial of the third kind

$$\hat{V}_n(x)=\frac{\cos\left(\left(n+\frac12\right)\arccos\,x\right)}{2^n\cos\frac{\arccos\,x}{2}}=\frac1{2^n}\left(U_n(x)-U_{n-1}(x)\right)$$

where $U_n(x)=\frac{\sin((n+1)\arccos\,x)}{\sqrt{1-x^2}}$ is the usual Chebyshev polynomial of the second kind, and the last relationship is derived through the trigonometric identity

$$2\sin\frac{\theta}{2}\cos\left(\left(n+\frac12\right)\theta\right)=\sin((n+1)\theta)-\sin\,n\theta$$

The question then is essentially asking for the negative of the coefficient of the $x^2$ term of $\frac18(U_3(x)-U_2(x))$ (Vieta); we can generate the two polynomials using the definition given above, or through an appropriate recursion relation. We then have

$$\begin{align*}U_2(x)&=4x^2-1\\U_3(x)&=8x^3-4x\end{align*}$$

and we thus have

$$\hat{V}_3(x)=\frac18((8x^3-4x)-(4x^2-1))=x^3-\frac{x^2}{2}-\frac{x}{2}+\frac18$$

which yields the identities

$$\begin{align*} \cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}&=\frac12\\ \cos\frac{\pi}{7}\cos\frac{3\pi}{7}+\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}+\cos\frac{\pi}{7}\cos\frac{5\pi}{7}&=-\frac12\\ \cos\frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}&=-\frac18 \end{align*}$$

Geometric picture: if $N$ is an integer greater than 1, the points $\left(\cos\frac{2\pi j}{N},\sin\frac{2\pi j}{N}\right)$, for $0\le j\lt N$, are equally distributed on the unit circle. More precisely, this set of points is unchanged if you rotate the plane by $\frac{2\pi}{N}$ about the origin. If we think of the points as a set of equal masses, then the center of mass has to be a point that is unchanged under rotation. There is only one such point.

With a little bit of work, the sum you have can be connected with this center of mass calculation.

To elaborate on the above: The matrix that rotates a vector by $\frac{2\pi}{N}$ is $$ R=\begin{bmatrix} \cos\frac{2\pi}{N} & -\sin\frac{2\pi}{N}\\\sin\frac{2\pi}{N} & \cos\frac{2\pi}{N} \end{bmatrix}. $$ Try multiplying $R$ by any of the vectors $\left(\cos\frac{2\pi j}{N},\sin\frac{2\pi j}{N}\right)$ and then applying addition formulas. You should get $\left(\cos\frac{2\pi (j+1)}{N},\sin\frac{2\pi (j+1)}{N}\right)$. But this means that the sum $$ S=\sum_{j=0}^{N-1}\left(\cos\frac{2\pi j}{N},\sin\frac{2\pi j}{N}\right) $$ is unchanged under multiplication by $R$. Can you prove that this implies $S=(0,0)$?

You are trying to prove $$ \frac{1}{2}-\cos\frac{\pi}{7}-\cos\frac{3\pi}{7}-\cos\frac{5\pi}{7}=0. $$ To relate it to the above, multiply both sides by 2, and use $\cos\theta=-\cos(\pi+\theta)=-\cos(\pi-\theta)$ to rewrite the left hand side.

Let $y = \cos\theta + i \cdot \sin\theta$, where $\theta$ has either of the values $$\frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \pi, \frac{9\pi}{7}, \frac{11\pi}{7} \ \text{and} \ \frac{13\pi}{7}$$ Then $y^{7} = \cos{7\theta} + i \:\sin{7\theta} =-1$. Then you have $$y^{7}+1=0 \Rightarrow (y+1)(y^{6}-y^{5}+y^{4}-y^{3}+y^{2}-y +1)=0$$ Now the root $y=-1$ corresponds to the value $\theta=\pi$. The roots of the equation $$y^{6}-y^{5}+y^{4}-y^{3}+y^{2}-y +1 =0 \qquad \cdots (1)$$ are therefore $\cos\theta + i\: \sin\theta$ where $\theta$ takes other values than $\pi$. Put $2x = y + \frac{1}{y} = 2 \cos\theta$ and note that:

• $\displaystyle y^{2}+\frac{1}{y^2} = 4x^{2}-2$

• $\displaystyle y^{3} + \frac{1}{y^3} = 8x^{3}-6x$.

Now divide $(1)$ by $y^{3}$ and get the answer.

Another method to look at this:

$7 \theta = \text{an odd multiple of} \: \pi$, therefore $\cos{4\theta} = -\cos{3\theta}$. Take $c=\cos{\theta}$. From this you have \begin{align*} 2\cdot (2c^{2}-1)^{2} -1 &= -(4c^{2}-3c) \\\ 8c^{4} + 4c^{2}-8c^{2} -3c + 1 &=0 \\\ (c+1) \cdot (8c^{3} - 4c^{2} -4c+1) &=0 \end{align*}

The following page also may be helpful: http://www.isibang.ac.in/~sury/luckyoct10.pdf