trigonometry - Resolve $A=cos{(pi/7)}+cos{(3pi/7)}+cos{(5pi/7)}$ using $u=A+iB$

With these two sums:

$$A=\cos(\pi/7)+\cos(3\pi/7)+\cos(5\pi/7)$$
$$B=\sin(\pi/7)+\sin(3\pi/7)+\sin(5\pi/7)$$

How to find the explicit value of $A$ using:

• $u=A+iB$


• the sum of $n$ terms in a geometric sequence: $u_0*\frac{1-q^{n+1}}{1-q}$

I know the answer is $\frac 12$ from this post, but there is no mention of this method.

Answer

Using Euler formula,

setting $2y=i\frac{\pi}7\implies e^{14y}=-1$

$$A+iB=\sum_{r=0}^2e^{(2r+1)2y}=e^{2y}\cdot\dfrac{1-(e^{4y})^3}{1-e^{4y}}=\dfrac{e^{2y}+1}{1-e^{4y}}=\dfrac1{1-e^{2y}}$$

Now, $\displaystyle\dfrac1{1-e^{i2u}}=\dfrac{-e^{-iu}}{e^{iu}-e^{-iu}}=-\dfrac{\cos u-i\sin u}{2i\sin u}=\dfrac12+i\cdot\dfrac{\cot u}2$

Now equate the real parts.