value of Trignometric expression

. \begin{align*} &\textup{Lets use complex numbers here}\\ & Let Z = cos(\frac{pi}{7})+isin(\frac{pi}{7})) \\ & \implies z^7=-1\\ & \implies (z+1)(z^6-z^5+z^4-z^3+z^2-z+1)=0\\ & \implies z^6-z^5+z^4-z^3+z^2-z+1=0\\ & \implies z+z^3+z^5=1+z^2+z^4+z^6\\ & \textup{(i)equating the real part gives the answer to the first problem.}\\ & \textup{equating imaginary part we get answer of the second problem}\\ &\implies \boxed{cos(\frac{pi}{7})-cos(\frac{2pi}{7})+cos(\frac{3pi}{7}) = \frac{1}{2}}\\ & \implies \boxed{sin(blah-blah)= 0}\\ & \end{align*}

Last problem is unsolvable i think..please see

http://www.wolframalpha.com/input/?i=sin%28pi%2F7%29+-+sin%282pi%2F7%29+%2B+sin%283pi%2F7%29&asynchronous=false&equal=Submit

Alternative for the first one(ok not really :p):

\epsilon_n = e^{\frac{(2n+1)\pi i}{7}}, n\in \left \{ -4,-3,\cdots, 2 \right \}
are 7 real roots of z^7 +1 = 0. So by vieta's relations ,

-1 + 2\left ( \cos\left ( \frac{\pi}{7} \right )+ \cos\left ( \frac{3\pi}{7} \right )+ \cos\left ( \frac{5\pi}{7} \right ) \right )= 0

oh yeah,
for the last one you get 0 = 0.
sorry.