Identities for some $\frac{k\pi}{7}$ and $\frac{k\pi}{14}$ angles

           The problem has been discussed here before. 

          I showed that $\left \{ cos\frac{\pi}{7},\;-cos\frac{2\pi}{7},\;cos\frac{3\pi}{7} \right \}=\left \{ sin\frac{5\pi}{14},\; -sin \frac{3\pi}{14},\; \sin\frac{\pi}{14} \right \}$ 

are the roots of the equation  $8x^3-4x^2-4x+1=0$ (see Remark 2, relations (5)-(8) ). And identities are expressions of  Vieta's relationships for this equation.

          For 3.i), let $A=cos\frac{\pi}{7}-cos \frac{2\pi}{7}+cos \frac{3\pi}{7}$. 

Let's calculate  $2sin\frac{\pi}{7} \cdot A=2sin\frac{\pi}{7}\cdot cos\frac{\pi}{7}-2sin\frac{\pi}{7}\cdot cos \frac{2\pi}{7}+2sin \frac{\pi}{7}\cdot cos \frac{5\pi}{7};$

with the formula of the double angle and the sum product formulas we get

$$2sin\frac{\pi}{7} \cdot A= sin\frac{2\pi}{7}-\left (sin \frac{3\pi}{7}-sin\frac{\pi}{7} \right )+ \left ( sin \frac{4\pi}{7}-sin\frac{2\pi}{7} \right ).$$

But the angles $\frac{4\pi}{7}$ and $\frac{3\pi}{7}$ are supplementary, so $sin \frac{4\pi}{7}=sin\frac{3\pi}{7}$, and we get $2sin\frac{\pi}{7} \cdot A= sin\frac{\pi}{7}$ from which 3.i) results.

          For 3.ii), let's note that $\frac{\pi}{2}-\frac{\pi}{7}=\frac{5\pi}{14},\;\frac {\pi}{2}-\frac{2\pi}{7}=\frac{3\pi}{14},\;\frac{\pi}{2}-\frac{3\pi}{7}=\frac{\pi}{14}$ 

and with the complement formula this is obtained from 3.i).

          For 3.iii), let $B=8\cdot cos\frac{\pi}{7}\cdot cos\frac{2\pi}{7} \cdot cos \frac{3\pi}{7}.$

 We will calculate $sin\frac{\pi}{7}\cdot B$ applying several times the formula of the double angle and the formulas of the supplements.

$$sin\frac{\pi}{7}\cdot B=4\cdot \left ( 2sin \frac{\pi}{7} \cdot cos\frac{\pi}{7}\right )\cdot cos\frac{2\pi}{7}\cdot cos\frac{3\pi}{7}=4\cdot sin\frac{2\pi}{7}\cdot cos \frac{2\pi}{7} \cdot cos \frac{3\pi}{7}=$$

$$=2\cdot \left ( 2 \cdot sin\frac{2\pi}{7}\cdot cos \frac{2\pi}{7} \right ) \cdot cos \frac{3\pi}{7}=2\cdot sin\frac{4\pi}{7}\cdot cos\frac{3\pi}{7}=2\cdot sin\frac{3\pi}{7}\cdot cos \frac{3\pi}{7}=$$

$$=sin\frac{6\pi}{7}=sin\frac{\pi}{7}$$

hence $B=1.$

          Now 3.iv) is obtained from 3.iii) with complement's formulae.

$\blacksquare$