Binomial Summations #4 #Challenge

Let $\large\begin{aligned}S &= \sum_{r=0}^{n} \left[\binom nr a^rb^{n-r}\cos\left(r\widehat B - (n-r)\widehat A\right)\right]\\R&=\sum_{r=0}^{n} \left[\binom nr a^rb^{n-r}e^{ir\widehat B}e^{-i(n-r )\widehat A}\right]\end{aligned}$

So, equating real and imaginary parts, $\large\begin{aligned}S &= \Re R\\&=\Re\sum_{r=0}^{n} \left[\binom nr a^rb^{n-r}e^{ir\widehat B}e^{-i(n-r)\widehat A}\right]\\&=\Re\left[a(\cos \widehat B+i\sin \widehat B)+b(\cos \widehat A-i\sin \widehat A)\right]^{\!n}\end{aligned}$ Now, because $a\sin \widehat B = b\sin \widehat A$ and also Projection Rule says $c = a\cos \widehat B + b\cos \widehat A$ , $\large\begin{aligned}\Re\left[a(\cos \widehat B+i\sin \widehat B)+b(\cos \widehat A-i\sin \widehat A)\right]^{\!n}&=\Re\left[a\cos \widehat B+b\cos \widehat A\right]^{\!n}\\&=\left(a\cos \widehat B+b\cos \widehat A\right)^{n}\\&=c^{n} \end{aligned}$

$\Large \displaystyle\therefore\sum_{r=0}^{n} \left[\binom {n}{r} a^rb^{n-r}\cos\left(r\widehat B - (n-r)\widehat A\right)\right] = c^{n}$