Trigonometry/The summation of finite series

Problem Statement
Find a closed form for
 * $$\sin(a)+\sin(a+b)+\sin(a+2b)+\cdots+\sin(a+(n-1)b)$$

Note: A 'closed form' is not mathematically defined, but just means a simplified formula which does not involve '...', or a summation sign. In our problem, we should look for a formula that only involves variables $$a,b,n$$, and known operations like the four operations, radicals, exponents, logarithm, and trigonometric functions.

Method 1
To sum the series
 * $$\sin(a)+\sin(a+b)+\sin(a+2b)+\cdots+\sin(a+(n-1)b)=S$$

Multiply each term by
 * $$2\sin\left(\tfrac{b}{2}\right)$$

Then we have
 * $$2\sin(a)\sin\left(\tfrac{b}{2}\right)=\cos\left(a-\tfrac{b}{2}\right)-\cos\left(a+\tfrac{b}{2}\right)$$

and similarly for all terms to
 * $$2\sin\bigl(a+(n-1)b\bigr)\sin\left(\tfrac{b}{2}\right)=\cos\left(a+(2n-3)\tfrac{b}{2}\right)-\cos\left(a+(2n-1)\tfrac{b}{2}\right)$$

Summing, we find that nearly all the terms cancel out and we are left with
 * $$2S\sin\left(\tfrac{b}{2}\right)=\cos\left(a-\tfrac{b}{2}\right)-\cos\left(a+(2n-1)\tfrac{b}{2}\right)=2\sin\left(a+(n-1)\tfrac{b}{2}\right)\sin\left(\tfrac{nb}{2}\right)$$

Hence
 * $$S=\sin\left(a+(n-1)\tfrac{b}{2}\right)\frac{\sin\left(n\tfrac{b}{2}\right)}{\sin\left(\tfrac{b}{2}\right)}$$

Similarly, if
 * $$C=\cos(a)+\cos(a+b)+\cos(a+2b)+\cdots+\cos\bigl(a+(n-1)b\bigr)$$

then
 * $$C=\cos\left(a+(n-1)\tfrac{b}{2}\right)\frac{\sin\left(n\tfrac{b}{2}\right)}{\sin\left(\tfrac{b}{2}\right)}$$

Method 2
Consider the following sum
 * $$s=e^{ai}+e^{(a+b)i}+\cdots+e^{(a+(n-1)b)i}$$

Since $$s$$ is a geometric series with common ratio $$e^{bi}$$, we get
 * $$s=\frac{e^{ai}(e^{nbi}-1)}{e^{bi}-1}=\frac{e^{ai}(e^{nbi}-1)}{e^{bi}-1}=\frac{e^{ai}e^\frac{nbi}{2}(e^\frac{nbi}{2}-e^{-\frac{nbi}{2}})}{e^\frac{bi}{2}(e^\frac{bi}{2}-e^{-\frac{bi}{2}})}$$
 * $$s=\frac{\sin\left(n\tfrac{b}{2}\right)}{\sin\left(\tfrac{b}{2}\right)}\cdot e^{\left(a+(n-1)\tfrac{b}{2}\right)i}$$

Therefore,
 * $$\sin(a)+\sin(a+b)+\sin(a+2b)+\cdots+\sin\bigl(a+(n-1)b\bigr)={\rm Im}(s)=\sin\left(a+(n-1)\tfrac{b}{2}\right)\frac{\sin\left(n\tfrac{b}{2}\right)}{\sin\left(\tfrac{b}{2}\right)}$$
 * $$\cos(a)+\cos(a+b)+\cos(a+2b)+\cdots+\cos\bigl(a+(n-1)b\bigr)={\rm Re}(s)=\cos\left(a+(n-1)\tfrac{b}{2}\right)\frac{\sin\left(n\tfrac{b}{2}\right)}{\sin\left(\tfrac{b}{2}\right)}$$