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

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- Mar 24th 2011, 02:14 PMbeczprove that if sin(a/2) ≠ 0 then...
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 - Mar 30th 2011, 10:16 AMveileen
$\displaystyle cosa + cos2a +...+cos(n+1)a=S$

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 $\displaystyle cos(\frac{r}{2})$. - Mar 31st 2011, 03:03 AMbecz
Not sure what you mean.

- Mar 31st 2011, 10:35 AMveileen
$\displaystyle cosa + cos2a +...+cos(n+1)a=S \Rightarrow (cosa + cos2a +...+cos(n+1)a)sin(\frac{a}{2})=S\cdot sin(\frac{a}{2})$

$\displaystyle sin a \cdot cos b=\frac{1}{2}(sin (a+b)+sin(a-b))$

"Is signs alternate multiply the sum by $\displaystyle cos(\frac{a}{2})$." for example: $\displaystyle S=cos a - cos 2a + cos 3a - cos 4a+...+(-1)^n cos(n+1)a$ - Apr 2nd 2011, 11:30 AMbecz
still not sure what you mean

- Apr 2nd 2011, 11:31 AMveileen
Okay, what do you think I mean?