Finding value of product of Cosines 
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$

My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\sin \frac{\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
So we have $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
Could some help me to solve it, Thanks in Advanced

Durgesh: to complement Jack D'Aurizio fantastic answer above.
If you use Euler's identity
$$e^{ix} = cos x + i\cdot sin x$$
and then apply for $-x$
$$e^{-ix} = cos (-x) + i\cdot  sin(-x) = cos x - i\cdot  sinx$$
and sum both equations, you get the well known
$$cos x={e^{ix}+e^{-ix}\over2}$$
so
$$\frac12+cos x={1+e^{ix}+e^{-ix}\over2}$$
that is
$$\frac12+cos x={1+e^{ix}+e^{2ix}\over{2e^{ix}}}={1\over{2e^{ix}}}{{e^{3ix}-1}\over{e^{ix}-1}}$$
Substituting for $3x$, $9x$ and $27x$ and multiplying all 4 equations, you get Jack's magical leap :-)
