# 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}.

Mar 15, 2017

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

Taking $\left(2 r - 1\right) \frac{\pi}{8} = \theta$

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

$= \left\{{\left[\sin \theta\right]}^{4} + {\left[\cos \theta\right]}^{4}\right\} .$

$= \left\{{\left[{\sin}^{2} \theta + {\cos}^{2} \theta\right]}^{2} - 2 {\sin}^{2} \theta {\cos}^{2} \theta\right\}$

$= \left\{{1}^{2} - \frac{1}{2} {\left(2 \sin \theta \cos \theta\right)}^{2}\right\}$

$= \left\{{1}^{2} - \frac{1}{2} {\sin}^{2} 2 \theta\right\}$

$= \left\{1 - \frac{1}{2} {\sin}^{2} \left(2 r - 1\right) \frac{\pi}{4}\right\}$

 ={1-1/2sin^2 ((rpi)/2-pi/4))}

$= \left\{1 - \frac{1}{2} {\left(\pm \frac{1}{\sqrt{2}}\right)}^{2}\right\} \textcolor{red}{\text{*}}$

$= \left\{1 - \frac{1}{4}\right\} = \frac{3}{4}$

$\left[\textcolor{red}{\text{*"=>sin((rpi)/2-pi/4)=pmsin(pi/4) or pmcos(pi/4)" for " r in ZZ^"+}}\right]$
So

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

$= {\sum}_{r = 1}^{r = 4} \left\{\frac{3}{4}\right\} = 4 \times \frac{3}{4} = 3$