Question

Find the Value of

0 cos 0° + 1 cos 1° + 2 cos 2° + 3 cos 3° +  ...... + 358 cos 358° + 359 cos 359°
$=\Sigma_{k=0}^{359}\ k\ Cos\ k^0$

Answer 1

Let y = 0 cos 0° + 1 cos 1° + 2 cos 2° + ...... + 358 cos 358° + 359 cos 359°

$y=\Sigma_{k=0}^{359}\ k\ cos\ k^0\\\\=\Sigma_{k=0}^{179}\ k\ cos\ k^0+\Sigma_{k=180}^{359}\ k\ cos\ k^0\\\\Let\ k=180+x\\\\y=\Sigma_{k=0}^{179}\ k\ cos\ k^0+\Sigma_{x=0}^{179}\ (180+x)\ cos\ (180^0+x^0)\\\\=\Sigma_{k=0}^{179}\ k\ cos\ k^0-\Sigma_{x=0}^{179}\ (180+x)\ cos\ x^0\\\\=-180 *\Sigma_{0}^{179}\ cos\ x^0\\\\=-180*[\Sigma_{x=0}^{89}\ cosx^0+\Sigma_{90}^{179}\ cos\ x^0]\\\\Let\ z=180-x\\\\y=-180*[\Sigma_{x=0}^{89}\ cosx^0+\Sigma_{z=90}^{1}\ cos(180-z)]$

$y=-180*[Cos0^0+\Sigma_{x=1}^{89}\ cosx^0-\Sigma_{z=1}^{89}\ cos\ z^0-cos90^0]\\\\=-180*1=-180$