Prove that cos4π/8 + cos43π/8 + …

Ans. $cos^4{\pi \over 8}+cos^4{3\pi \over 8}+cos^4{4\pi \over 8}+cos^4{7\pi \over 8} ={3\over 2}$

Taking LHS,

$cos^4{\pi \over 8}+cos^4{3\pi \over 8}+cos^4{4\pi \over 8}+cos^4{7\pi \over 8}$

=> $cos^4{\pi \over 8}+cos^4{3\pi \over 8}+cos^4[{{\pi} -{3\pi \over 8}}]+cos^4[{{\pi - {\pi \over 8}}}]$

=> $cos^4{\pi \over 8}+cos^4{3\pi \over 8}+({-cos{3\pi \over 8}})^4+({-cos{\pi \over 8}})^4$

=> $2[cos^4{\pi \over 8}+cos^4{3\pi \over 8}]$

=> $2[cos^4{\pi \over 8}+cos^4({\pi \over 2}-{\pi \over 8})]$

=> $2[cos^4{\pi \over 8}+sin^4{\pi \over 8}]$

$=> 2[(cos^2{\pi \over 8}+sin^2{\pi \over 8})^2 -2 cos^2{\pi \over 8}.sin^2{\pi \over 8}]$

$=> 2[(1)^2 -{1\over 2}(2 cos{\pi \over 8}.sin{\pi \over 8})^2]$

$=> 2[1 -{1\over 2}(sin {\pi \over 4})^2]$

$=> 2[1 -{1\over 2}\times {1\over 2}]$

$=> 2[1 -{1\over 4}] => 2 \times {3\over 4} = {3\over 2} = RHS$

Hence Proved