Know some trigonometry

Let $ω=\cos\frac{2π}{7}+i\sin\frac{2π}{7}$ be the $7th$ root of unity.

Hence, $1+ω+ω^{2}+....+ω^{6}=0$ .Comparing the real parts, we have

$1+\cos\frac{2π}{7}+\cos\frac{4π}{7}+\cos\frac{6π}{7}+\cos\frac{8π}{7}+\cos\frac{10π}{7}+\cos\frac{12π}{7}=0$ .

Using $\cos\theta=-\cos(π-\theta)=-\cos(\theta-π)$ ,we have,

$2(\cos\frac{π}{7}-\cos\frac{2π}{7}+\cos\frac{3π}{7})=1$

For the benefit of some readers, I will write up a solution that does not explicitly refer to roots of unity.

Since the points $(\cos\frac{k\pi}{7},\sin\frac{k\pi}{7})$ , for odd $k$ , are equally spaced on the unit circle, forming a regular heptagon, the sum of their coordinates is 0, by symmetry(sketch this heptagon to better understand the solution!). Thus the sum of all $\cos\frac{k\pi}{7}$ is 0, where $k$ runs through all odd integers from 1 through 13. For $k=7$ we get $\cos{\pi}=-1$ . Now the sum for $k=9,11,13$ is the same as for $k=1,3,5$ , for example, $\cos\frac{11\pi}{7}=\cos\frac{3\pi}{7}$ (the heptagon is symmetric with respect to the x-axis). Thus $2\cos\frac{\pi}{7}+2\cos\frac{3\pi}{7}+2\cos\frac{5\pi}{7}-1=0$ . Since $\cos\frac{2\pi}{7}=-\cos\frac{5\pi}{7}$ , we have $\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$ $=\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}=\frac{1}{2}$ .

Solution using complex numbers.

We know that $cos\left( \frac { 2\pi }{ 7 } \right) =-cos\left( \frac { 5\pi }{ 7 } \right)$

Therefore, $S=cos\left( \frac { \pi }{ 7 } \right) +cos\left( \frac { 3\pi }{ 7 } \right) +cos\left( \frac { 5\pi }{ 7 } \right)$

Take the following complex number: $z=cis\left( \frac { \pi }{ 7 } \right)$

By De Moivre's formula: ${ z }^{ 7 }=cis\left( \pi \right) =-1$ .......... Equation 1

As z is not 1, we can factor this as:

${ z }^{ 6 }-{ z }^{ 5 }+{ z }^{ 4 }-{ z }^{ 3 }+{ z }^{ 2 }-{ z }+1=0$ .......... Equation 2

Using the relation between cis and cos:

$z+{ z }^{ -1 }=2cos\left( \frac { \pi }{ 7 } \right)$

${ z }^{ 3 }+{ z }^{ -3 }=2cos\left( \frac { 3\pi }{ 7 } \right)$

${ z }^{ 5 }+{ z }^{ -5 }=2cos\left( \frac { 5\pi }{ 7 } \right)$

Adding the expressions above:

${ z }+{ z }^{ -1 }+{ z }^{ 3 }+{ z }^{ -3 }+{ z }^{ 5 }+{ z }^{ -5 }=2S$

Using equation 1: ${ z }^{ -1 }=-{ z }^{ 6 }$ and ${ z }^{ -3 }=-{ z }^{ 4 }$ and ${ z }^{ -5 }=-{ z }^{ 2 }$

Therefore,

${ z }-{ z }^{ 6 }+{ z }^{ 3 }-{ z }^{ 4 }+{ z }^{ 5 }-{ z }^{ 2 }=2S$

Using equation 2: 2S=1

$\boxed{S=\frac { 1 }{ 2 }}$

use cos2x and cos3x expansions...that is ${ cos }^{ 2 }x-{ sin }^{ 2 }x$ and ${ 4cos }^{ 3 }x-3cosx$

cos (180/7) -cos(360/7) +cos (540/7) = 1/2 use pi = 180 ; then you can replace the values : 180x1=180
180x 2 = 360 180 x 3 = 540

cos (pi/7) - cos (2pi/7) + cos (3pi/7)= 2 cos (2pi/7) cos (pi/7) - cos (2pi/7) = cos (2pi/7) (2cos pi/7 - 1) = cos (2pi/7) (-2 cos 6pi/7 -1) = cos (2pi/7) (4 sin^{2} (3pi/7) - 3) = cos (2pi/7) [-sin (9pi/7)]/ sin (3pi/7) = cos (2pi/7) sin (2pi/7)/sin (3pi/7) = (1/2) (sin 4pi/7)/sin (3pi/7) = 1/2.