Fun with Trigonometry!!

$\left( 1\quad +\quad cos\frac { 5\pi }{ 8 } \right) = cos\left( \pi \quad -\quad \frac { 3\pi }{ 8 } \right) = -cos\frac { 3\pi }{ 8 }$

Similarly,

$\left( 1\quad +\quad cos\frac { 7\pi }{ 8 } \right) = -cos\frac { \pi }{ 8 }$

Therefore,

$\left( 1\quad +\quad cos\frac { \pi }{ 8 } \right) \left( 1\quad +\quad cos\frac { 3\pi }{ 8 } \right) \left( 1\quad -\quad cos\frac { 3\pi }{ 8 } \right) \left( 1\quad -\quad cos\frac { \pi }{ 8 } \right)$

= $\left( 1\quad -\quad { cos }^{ 2 }\frac { 3\pi }{ 8 } \right) \left( 1\quad -\quad { cos }^{ 2 }\frac { \pi }{ 8 } \right)$

Now, we know that cos²(x) = (1 + cos(2x))/2

cos²(π/8) = (1 + cos(π/4))/2 = (1 + √2/2)/2 = 1/2 + √2/4 = (2 + √2)/4

cos²(3π/8) = - (2 + √2)/4

= $\left( 1\quad -\quad \frac { 2\quad +\quad \sqrt { 2 } }{ 4 } \right) \left( 1\quad +\quad \frac { 2\quad +\quad \sqrt { 2 } }{ 4 } \right)$

= $\frac { 4\quad -\quad 2 }{ 16 } = 1/8$

Hence, x + y = 9

We know cos 2x =2(cos^2 x) -1.so 1+ cos 2x =2 cos^2 x...using this we write (1+cos pie/8 )= 2 cos^2 pie/16 , (1+cos 3pie/8 )= 2 cos^2 3pie/16 , (1+cos 5pie/8 )= 2 cos^2 5pie/16 , (1+ cos 7pie/8 )= 2 cos^2 7pie/16 ...now sin (pie/2 - 7pie/16 ) =cos 7pie/16 ...so sin pie/16 = cos 7pie/16...sin (pie/2 - 5pie/16 ) = cos 5pie/16 ....so sin 3pie/16 = cos 5pie/16...

now answer=2 cos^2 pie/16 * 2 cos^2 3pie/16 * 2 cos^2 5pie/16 * 2 cos^2 7pie/16... =2 cos^2 pie/16 2 cos^2 3pie/16 * 2 sin^2 3 pie/16 * 2 sin^2 pie/16... =(4 cos^2 pie/16 * sin^2 pie/16) * (4 cos^2 3pie/16 * sin ^2 3pie/16)... =sin^2 pie/8 * sin^2 3pie/8 (using sin 2x = 2 sin x cos x)... now , cos(pie/2 - 3pie/8 )= sin 3pie/8....so cos pie/8 = sin 3pie/8.... therefore answer = sin^2 pie/8 * cos^2 pie/8.... =(1/4) (4 sin^2 pie/8 * cos^2 pie/8)... =(1/4)* sin^2 pie/4... =(1/4)*(1/2)=1/8...