Thank you for reporting, we will resolve it shortly
Q.
$\displaystyle\sum_{r=0}^{n}{ }^{n} C _{r} \sin r x \cos (n-r) x=$
Binomial Theorem
Solution:
We have, $ \displaystyle\sum_{r=0}^{n} C_{r} \sin r x \cos (n-r) x$
$= \frac{1}{2}\left[\left(n C_{0} \sin 0 x \cos n x+n C n \sin n x \cos 0 x\right)\right.$
$+\left(n C_{1} \sin x \cos (n-1) x+n C n_{-1} \sin (n-1) x \cdot \cos x\right)$
$+\left(n C_{2} \sin 2 x \cos (n-2) x+n C n_{-2} \sin (n-2) x \cdot \cos 2 x\right) $ $\left.+\ldots+\left(n C n \sin n x \cos 0 x+n C_{0} \sin 0 x \cos n x\right)\right] $
$= \frac{1}{2}\left[n C_{0} \sin n x+n C_{1} \sin n x+\ldots+n C n \sin n x\right] $
$= \frac{1}{2}\left[n C_{0}+n C_{1}+\ldots+n C n\right] \sin n x=\frac{2^{n} \sin n x}{2} $
$ \therefore \displaystyle\sum_{r=0}^{n} C_{r} \sin r x \cos (n-r) x=2 n-1 \sin n x$.