Let $z=\cos{x} + i\sin{x}$, then $$ \cos{kx}=\frac{1}{2}\times\Big(z^k + \frac{1}{z^k}\Big)\quad\text{and}\quad\sin{kx}=-\frac{1}{2}\times\Big(z^k - \frac{1}{z^k}\Big)i $$ It follows that \begin{align*} &\cos{x} + \cos{2x} + \cdots + \cos{nx}\\ =&\frac{1}{2}\times\Big(z + \frac{1}{z}\Big)+\frac{1}{2}\times\Big(z^2 + \frac{1}{z^2}\Big)+\cdots+\frac{1}{2}\times\Big(z^n + \frac{1}{z^n}\Big)\\ =&\frac{1}{2}\Big(z+z^2+\cdots+z^n\Big)+\frac{1}{2}\Big(\frac{1}{z}+\frac{1}{z^2}+\cdots +\frac{1}{z^n}\Big)\\ =&\frac{1}{2}\times z\times\frac{1-z^{n}}{1-z}+\frac{1}{2}\times\frac{1}{z}\times\frac{1-\frac{1}{z^{n}}}{1-\frac{1}{z}}\\ =&\frac{1}{2}\times\frac{(z^n-1)(z^{n+1}+1)}{z^n(z-1)} \end{align*} In order to convert it back to trigonometric expression, we need to transform every term to the form of $(x^k\pm\frac{1}{x^k})$. Dividing both the denominator and numerator part by $z{n+\frac{1}{2}}$ yields $$\frac{1}{2}\times\frac{(z^{\frac{n}{2}}-\frac{1}{z^{\frac{n}{2}}})(z^{\frac{n+1}{2}}+\frac{1}{z^{\frac{n+1}{2}}})}{z^\frac{1}{2}-\frac{1}{z^\frac{1}{2}}}=\boxed{\frac{\sin(\frac{nx}{2})\cos\Big(\frac{(n+1)x}{2}\Big)}{\sin(\frac{x}{2})}}$$ Using a similar method, we have \begin{align*} &\sin{x} + \sin{2x} + \cdots + \sin{nx}\\ =&-\frac{i}{2}\Big[\Big(z-\frac{1}{z}\Big)+\Big(z^2-\frac{1}{z^2}\Big)+\cdots+\Big(z^n-\frac{1}{z^n}\Big)\Big]\\ =&-\frac{i}{2}\Big[\Big(z+z^2+\cdots+z^n\Big)-(\frac{1}{z}+\frac{1}{z^2}+\cdots+\frac{1}{z^n}\Big)\Big]\\ =&-\frac{i}{2}\times\Big[ z\times\frac{1-z^{n}}{1-z}-\frac{1}{z}\times\frac{1-\frac{1}{z^{n}}}{1-\frac{1}{z}}\Big]\\ =&-\frac{i}{2}\times\frac{(z^n-1)(z^{n+1}-1)}{z^n(z-1)}\\ =&-\frac{i}{2}\times\frac{(z^{\frac{n}{2}}-\frac{1}{z^{\frac{n}{2}}})(z^{\frac{n+1}{2}}-\frac{1}{z^{\frac{n+1}{2}}})}{z^\frac{1}{2}-\frac{1}{z^\frac{1}{2}}}\\ =&\boxed{\frac{\sin(\frac{nx}{2})\sin\Big(\frac{(n+1)x}{2}\Big)}{\sin(\frac{x}{2})}} \end{align*}