prove that if sin(a/2) ≠ 0 then
cosa + cos2a +...+cos(n+1)a = (sin((n+1)a/2)*cos((n+1)a/2)/sin(a/2)
FInd this sum also in the case sin(a/2) = 0
Find a similar formula for sina + sin 2a +...+ sin(n+1)a
$\displaystyle cosa + cos2a +...+cos(n+1)a=S$
Multiply it by $\displaystyle sin(\frac{a}{2})$. When you have sums like this - multiply by $\displaystyle sin(\frac{r}{2})$, r is ratio. Is signs alternate multiply the sum by $\displaystyle cos(\frac{r}{2})$.
$\displaystyle cosa + cos2a +...+cos(n+1)a=S \Rightarrow (cosa + cos2a +...+cos(n+1)a)sin(\frac{a}{2})=S\cdot sin(\frac{a}{2})$
$\displaystyle sin a \cdot cos b=\frac{1}{2}(sin (a+b)+sin(a-b))$
"Is signs alternate multiply the sum by $\displaystyle cos(\frac{a}{2})$." for example: $\displaystyle S=cos a - cos 2a + cos 3a - cos 4a+...+(-1)^n cos(n+1)a$