# If sin18@=(sqrt(5)-1)/4 and cos36@=(sqrt(5)+1)/4 then prove that (1+cos(pi/10))(1+cos((3pi)/10))(1+cos((7pi)/10))(1+cos((9pi)/10))=(1/16)?

Oct 23, 2017

$L H S = \left(1 + \cos \left(\frac{\pi}{10}\right)\right) \left(1 + \cos \left(\frac{3 \pi}{10}\right)\right) \left(1 + \cos \left(\frac{7 \pi}{10}\right)\right) \left(1 + \cos \left(\frac{9 \pi}{10}\right)\right)$

Puttingv$\frac{\pi}{10} = 2 \theta$ we have

$L H S = \left(1 + \cos \left(2 \theta\right)\right) \left(1 + \cos \left(6 \theta\right)\right) \left(1 + \cos \left(14 \theta\right)\right) \left(1 + \cos \left(18 \theta\right)\right)$

$= \left(2 {\cos}^{2} \left(\theta\right)\right) \left(2 {\cos}^{2} \left(3 \theta\right)\right) \left(2 {\cos}^{2} \left(7 \theta\right)\right) \left(2 {\cos}^{2} \left(9 \theta\right)\right)$

$= {\left[\left(2 \cos \left(\theta\right) \cos \left(9 \theta\right)\right) \left(2 \cos \left(3 \theta\right) \cos \left(7 \theta\right)\right)\right]}^{2}$

$= {\left[\left(\cos \left(10 \theta\right) + \cos \left(8 \theta\right)\right) \left(\cos \left(10 \theta\right) + \cos \left(4 \theta\right)\right)\right]}^{2}$

$= {\left[\left(\cos \left(\frac{\pi}{2}\right) + \cos \left(8 \theta\right)\right) \left(\cos \left(\frac{\pi}{2}\right) + \cos \left(4 \theta\right)\right)\right]}^{2}$

$= {\left[\cos \left(8 \theta\right) \cos \left(4 \theta\right)\right]}^{2}$

$= {\left[\frac{4 \cos \left(8 \theta\right) \cos \left(4 \theta\right) \sin \left(4 \theta\right)}{4 \sin \left(4 \theta\right)}\right]}^{2}$

$= \frac{1}{16} {\left[\sin \frac{16 \theta}{\sin} \left(4 \theta\right)\right]}^{2}$

$= \frac{1}{16} {\left[\sin \frac{\frac{8 \pi}{10}}{\sin} \left(\frac{2 \pi}{10}\right)\right]}^{2}$

$= \frac{1}{16} {\left[\sin \frac{\pi - \frac{2 \pi}{10}}{\sin} \left(\frac{2 \pi}{10}\right)\right]}^{2}$

$= \frac{1}{16} {\left[\sin \frac{\frac{2 \pi}{10}}{\sin} \left(\frac{2 \pi}{10}\right)\right]}^{2}$

$= \left(\frac{1}{16}\right) = R H S$