Find the value of sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}.?

sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}.

1 Answer
Mar 15, 2017

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