Sum of sines.

Relevant wiki: De Moivre's Theorem

For the main problem

Consider the following sum

$\begin{aligned} \displaystyle \sum_{k=0}^n \dbinom{n}{k} \left[ \cos (kx) \pm i \sin(kx) \right] &= \sum_{k=0}^n \dbinom{n}{k} e^{\pm ikx} \\ &= \sum_{k=0}^n \dbinom{n}{k} {\left( e^{\pm ix} \right)}^k \\ &= {\left( 1 + e^{\pm ix} \right)}^n \end{aligned}$

Hence

$\begin{aligned} \displaystyle \sum_{k=0}^n \dbinom{n}{k} \sin(kx) &= \dfrac{ \displaystyle \left\{ \sum_{k=0}^n \dbinom{n}{k} \left[ \cos (kx) + i \sin(kx) \right] \right\} - \left\{ \sum_{k=0}^n \dbinom{n}{k} \left[ \cos (kx) - i \sin(kx) \right] \right\} }{2i} \\ \\ &= \dfrac{ {\left( 1 + e^{ix} \right)}^n - {\left( 1 + e^{- ix} \right)}^n }{2i} \\ \\ &= \dfrac{ {\left[ 1 + \cos(x) + i \sin(x) \right]}^n - {\left[ 1 + \cos(x) - i \sin(x) \right]}^n }{2i} \\ \\ &= \dfrac{ {\left[ 2 \cos^2 \left( \frac x2 \right) + 2i \sin \left( \frac x2 \right) \cos \left( \frac x2 \right) \right]}^n - {\left[ 2 \cos^2 \left( \frac x2 \right) - 2i \sin \left( \frac x2 \right) \cos \left( \frac x2 \right) \right]}^n }{2i} \\ \\ &= \dfrac{ 2^n \cos^n \left( \frac x2 \right) \left\{ {\left[ \cos \left( \frac x2 \right) + i \sin \left( \frac x2 \right) \right]}^n - {\left[ \cos \left( \frac x2 \right) - i \sin \left( \frac x2 \right) \right]}^n \right\} }{2i} \\ \\ &= \dfrac{ 2^n \cos^n \left( \frac x2 \right) \left\{ {\left( e^{\frac{ix}{2}} \right)}^n - {\left( e^{\frac{-ix}{2}} \right)}^n \right\} }{2i} \\ \\ &= 2^n \cos^n \left( \dfrac x2 \right) \left[ \dfrac{ {\left( e^{\frac{inx}{2}} \right)} - {\left( e^{\frac{-inx}{2}} \right)} }{2i} \right] \\ \\ &= \boxed{2^n \cos^n \left( \dfrac x2 \right) \sin \left( \dfrac{nx}{2} \right)} \end{aligned}$

For the bonus problem

$\begin{aligned} \displaystyle \sum_{k=0}^n \dbinom{n}{k} \cos(kx) &= \dfrac{ \displaystyle \left\{ \sum_{k=0}^n \dbinom{n}{k} \left[ \cos (kx) + i \sin(kx) \right] \right\} + \left\{ \sum_{k=0}^n \dbinom{n}{k} \left[ \cos (kx) - i \sin(kx) \right] \right\} }{2} \\ \\ &= \dfrac{ {\left( 1 + e^{ix} \right)}^n + {\left( 1 + e^{- ix} \right)}^n }{2} \\ \\ &= \dfrac{ {\left[ 1 + \cos(x) + i \sin(x) \right]}^n + {\left[ 1 + \cos(x) - i \sin(x) \right]}^n }{2} \\ \\ &= \dfrac{ {\left[ 2 \cos^2 \left( \frac x2 \right) + 2i \sin \left( \frac x2 \right) \cos \left( \frac x2 \right) \right]}^n + {\left[ 2 \cos^2 \left( \frac x2 \right) - 2i \sin \left( \frac x2 \right) \cos \left( \frac x2 \right) \right]}^n }{2} \\ \\ &= \dfrac{ 2^n \cos^n \left( \frac x2 \right) \left\{ {\left[ \cos \left( \frac x2 \right) + i \sin \left( \frac x2 \right) \right]}^n + {\left[ \cos \left( \frac x2 \right) - i \sin \left( \frac x2 \right) \right]}^n \right\} }{2} \\ \\ &= \dfrac{ 2^n \cos^n \left( \frac x2 \right) \left\{ {\left( e^{\frac{ix}{2}} \right)}^n + {\left( e^{\frac{-ix}{2}} \right)}^n \right\} }{2} \\ \\ &= 2^n \cos^n \left( \dfrac x2 \right) \left[ \dfrac{ {\left( e^{\frac{inx}{2}} \right)} + {\left( e^{\frac{-inx}{2}} \right)} }{2} \right] \\ \\ &= \boxed{2^n \cos^n \left( \dfrac x2 \right) \cos \left( \dfrac{nx}{2} \right)} \end{aligned}$

Relevant wiki: Euler's Formula

$\begin{aligned} S & = \sum_{k=0}^n {n \choose k} \sin kx \\ & = \sum_{k=0}^n {n \choose k} \Im \left \{e^{ikx} \right \} & \small \color{#3D99F6} \text{By Euler's formula: } e^{i\theta} = \cos \theta + i \sin \theta \\ & = \Im \left \{\sum_{k=0}^n {n \choose k} e^{ikx} \right \} & \small \color{#3D99F6} \text{where }\Im\{z\} \text{ is the real part of complex number } z. \\ & = \Im \left \{\left(1+e^{ix} \right)^n \right \} \\ & = \Im \left \{\left(1+\cos x + i\sin x \right)^n \right \} \\ & = \Im \left \{\left(1+2\cos^2 \frac x2 - 1 + 2 i \sin \frac x2 \cos \frac x2 \right)^n \right \} \\ & = \Im \left \{2^n \cos^n \frac x2 \left( \cos \frac x2 + i \sin \frac x2 \right)^n \right \} \\ & = \Im \left \{2^n \cos^n \frac x2 \cdot e^{i \frac {nx} 2} \right \} \\ & = \boxed {2^n \cos^n \dfrac x2 \sin \dfrac {nx} 2} \end{aligned}$

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Bonus : Similar solution see here .

For this problem, two useful tools are:

Binomial Theorem: $(a+b)^n = \displaystyle \sum_{k=0}^n a^k b^{n-k}$ ;

Euler's formula: $e^{ix} = cos(x) + i \text{ } sin(x) \text{ } \Rightarrow \text{ } sin(x) = \frac{1}{2i} (e^{ix} - e^{-ix})$ and $cos(x) = \frac{1}{2} (e^{ix} + e^{-ix})$ .

Then,

$\displaystyle \sum_{k=0}^n {n \choose k} sin(kx) = \sum_{k=0}^n {n \choose k} \frac{1}{2i} (e^{ikx} - e^{-ikx})$

$\displaystyle = \frac{1}{2i} \sum_{k=0}^n {n \choose k} e^{ikx} - \frac{1}{2i} \sum_{k=0}^n {n \choose k} e^{-ikx}$

$\displaystyle = \frac{1}{2i} (e^{ix}+1)^n - \frac{1}{2i} (e^{-ix}+1)^n \qquad$ [Binomial Thm with $a=e^{\pm ix}$ and $b=1$ ]

$\displaystyle = \frac{1}{2i} (e^{ix/2})^n(e^{ix/2}+e^{-ix/2})^n - \frac{1}{2i} (e^{-ix/2})^n(e^{-ix/2}+e^{ix/2})^n$

$\displaystyle = \frac{1}{2i} (e^{inx/2} - e^{-inx/2}) (e^{ix/2}+e^{-ix/2})^n$

$\displaystyle = sin(nx/2)(2 \; cos(x/2))^n \qquad$ [from the inverted Euler's formula for $sin(nx)$ ]

$\Rightarrow \boxed{\displaystyle \sum_{k=0}^n {n \choose k} sin(kx) = 2^n cos^n(x/2) sin(nx/2)}$ .

For the bonus problem, observe that $cos(x) = \frac{1}{2} (e^{ix} + e^{-ix})$ and that, in the above derivation, the factor $\frac{1}{2i}$ and the $-$ sign between the $e^{\pm ikx}$ terms from the $sin(kx)$ replacement in the first equality are the same as those in the next-to-last equality. So, the replacement $\frac{1}{2i} \rightarrow \frac{1}{2}$ and $- \rightarrow +$ , respectively, of those in the derivation and using the inverted Euler's formula for $cos(x)$ immediately yields:

$\Rightarrow \boxed{\displaystyle \sum_{k=0}^n {n \choose k} cos(kx) = 2^n cos^n(x/2) cos(nx/2)}$ .