Question

If $S_1={\sum }_{k=0}^n{{}^{n}C}_k\cos \left(kx\right)\cos (n-k)x$ and $S_2={\sum }_{k=0}^n{{}^{n}C}_k\sin \left(kx\right)\sin (n-k)x$, then $S_1-S_2$ equals

• A. $2^{n-1}\cos \left(nx\right)$
• B. $2^n\cos \left(nx\right)$
• C. $0$
• D. None of these

Answer 1

The correct option is B $2^n\cos \left(nx\right)$

$S_1=\sum {}^n{C}_k\cos \left(kx\right)\cos (n-k)x$
$S_2=\sum {}^n{C}_k\sin (n-k)x\sin \left(k\right)x$
Hence $S_1-S_2$
$=S$
$=\sum {}^n{C}_k\left[\cos \right(kx\left)\cos \right((n-k)x)-\sin (kx\left)\sin \right((n-k)x\left)\right]$
$=\sum {}^n{C}_k\cos \left(nx\right)$
Hence
$S=2^n\cos \left(nx\right)$