sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}
Taking (2r-1)pi/8=theta
{[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}
= {[sintheta]^4 +[costheta]^4}.
={[sin^2theta +cos^2theta]^2-2sin^2thetacos^2theta}
={1^2-1/2(2sinthetacostheta)^2}
={1^2-1/2sin^2 2theta}
={1-1/2sin^2 (2r-1)pi/4}
={1-1/2sin^2 ((rpi)/2-pi/4))}
={1-1/2(pm1/sqrt2)^2}color(red)"*"
={1-1/4 }=3/4
[color(red)"*"=>sin((rpi)/2-pi/4)=pmsin(pi/4) or pmcos(pi/4)" for " r in ZZ^"+"]
So
sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}
=sum_(r=1)^(r=4) {3/4}=4xx3/4=3