Sum with Binomial Coefficients and Sine; $S=\sum_{k=0}^n \binom{n}{k} \sin(kx)$

Question

Sum with Binomial Coefficients

Let $n ∈ ℕ₀$ and $x ∈ ℝ$.

$$S=\sum_{k=0}^n \binom{n}{k} \sin(kx)$$

Simplify the sum to a polynomial in n.

I tried to use Euler's Formula and the Binomial Theorem, similar to Tricky Sum involving Binomial Coefficients and Sine. Alas I could not fully understand the option presented. I would be grateful for an explanation or even better another way of approaching the problem.

Answer 1

We have;

$$S=\sum_{k=0}^n \binom{n}{k} sin(kx)$$

I can think of two methods;

Method [1];

We can begin with $k=0,1,2,\cdots$ and consequently one can find a pattern and you will observe; $$S=2^nsin\left(\frac{nx}{2}\right)cos^n\left(\frac{x}{2}\right)$$

Method [2];

From Euler's Identity; $e^{ix}=isinx+cosx$, substitute $Sin(kx)= Im(e^{ikx})$, $$S=\sum_{k=0}^n \binom{n}{k} Im(e^{ikx})$$ $$S=Im \left(\sum_{k=0}^n \binom{n}{k} (e^{ix})^k\right)$$

$$S=Im \left(1+e^{ix}\right)^n$$

$$S=Im \left(isinx+cosx+1\right)^n$$

$$S=Im \left(isinx+2cos^2\frac{x}{2}\right)^n$$

$$S=Im \left(2isin\frac{x}{2}cos\frac{x}{2}+2cos^2\frac{x}{2}\right)^n$$

$$S=2^nIm \left(isin\frac{x}{2}cos\frac{x}{2}+cos^2\frac{x}{2}\right)^n$$

$$S=2^n\,Im \left(e^{\frac{inx}{2}}cos^n\frac{x}{2}\right)$$

Using De Moivre theorem, and you can conclude here with

$$\sum_{k=0}^n \binom{n}{k} sin(kx)=2^nsin\left(\frac{nx}{2}\right)cos^n\left(\frac{x}{2}\right)$$

Extras - $cos2x=2cos^2x-1, sin2x=2sinxcosx$