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

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

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Answer 1 · 2017-03-15T06:14:39

# 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#