Evaluate $\sum_{r=0}^n \binom{n}{r}\sin rx \cos (n-r)x$

Question

Evaluate

$$ \sum_{r=0}^n \left[\binom{n}{r}\cdot\sin rx \cdot \cos (n-r)x\right] $$

I tried to use binomial identities, but since there are trigonometric terms, I don't have the idea how to approach it.

Can anyone please help?

Answer 1

Hint:

Since $${n\choose r}={n\choose {n-r}}\qquad \text{and}\qquad \sin(a+b)=\sin a\cos b+\sin b\cos a$$ we have $${n\choose r}\sin rx\cdot\cos (n-r)x+{n\choose {n-r}}\sin(n-r)x\cdot \cos rx={n\choose r}\sin nx$$

Answer 2

Let,

$\text{S} =\displaystyle \sum_{r=0}^n \left[\dbinom{n}{r}\cdot\sin (rx) \cdot \cos (n-r)x\right]$

$=\dfrac{1}{2}\displaystyle \sum_{r=0}^n \left[\dbinom{n}{r}\cdot2\sin (rx) \cos (n-r)x\right]$

$= \dfrac{1}{2}\displaystyle \sum_{r=0}^n \dbinom{n}{r}\cdot (\sin(nx)+\sin(2r-n)x)$

$= 2^{n-1}\sin(nx)+\displaystyle\dfrac{1}{2}\sum_{r=0}^n \dbinom{n}{r}\cdot\sin(2r-n)x$

Now,

$\text{J}=\displaystyle\sum_{r=0}^n \dbinom{n}{r}\cdot\sin(2r-n)x $

$=\displaystyle\sum_{r=0}^n \dbinom{n}{n-r}\cdot\sin[2(n-r)-n]x $ $\left(\because \displaystyle\sum_{r=0}^n f(r) = \sum_{r=0}^n f(n-r)\right)$

$=-\displaystyle\sum_{r=0}^n \dbinom{n}{r}\cdot\sin(2r-n)x $

$\implies \text{J}=-\text{J}$

$\implies \text{J}=0$

$\therefore S=\boxed{2^{n-1}\sin(nx)}$

Answer 3

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Note that $\ds{\pars{a - \ol{a}}\pars{b + \ol{b}} = ab + a\ol{b} - \ol{a}b - \ol{a}\ol{b} =2\ic\,\Im\pars{ab + a\ol{b}}}$

\begin{align} &\color{#f00}{% \sum_{r = 0}^{n}{n \choose r}\sin\pars{rx}\cos\pars{\bracks{n - r}x}} = \sum_{r = 0}^{n}{n \choose r}{\expo{\ic rx} - \expo{-\ic rx} \over 2\ic}\, {\expo{\ic\pars{n - r}x} + \expo{\ic\pars{r - n}x} \over 2} \\[3mm] = &\ -\,{1 \over 4}\,\ic\pars{2\ic}\Im\sum_{r = 0}^{n}{n \choose r}\pars{% \expo{\ic n x} + \expo{-\ic nx}\expo{2\ic rx}} \\[3mm] = &\ \half\,\Im\bracks{\expo{\ic nx}\sum_{r = 0}^{n}{n \choose r} + \expo{-\ic nx}\sum_{r = 0}^{n}{n \choose r}\pars{\expo{2\ic x}}^{r}} = \half\,\Im\bracks{\expo{\ic nx}2^{n} + \expo{-\ic nx}\pars{1 + \expo{2\ic x}}^{n}} \\[3mm] = &\ \half\,\Im\bracks{\expo{\ic nx}2^{n} + 2^{n}\cos^{n}\pars{x}} = \color{#f00}{2^{n - 1}\sin\pars{nx}} \end{align}