What is the formula and how to derive the formula of sina+sin2a+sin3a...........?
Take a look at this Wikipedia article on trigonometric identities. The sum and difference formulas are in there.
For this problem, start writing out each term as a sum of the previous term and look for a pattern.
$\displaystyle \sin(a)$
$\displaystyle \sin(2a)=\sin(a)\cos(a)+ \sin(a)\cos(a)=2\sin(a)\cos(a)$
$\displaystyle \sin(3a)=\sin(2a+a)=\sin(2a)\cos(a)+\sin(a)\cos(2a )$
We already know the sin(2a) though and it isn't hard to find the cos(2a). Every next term can use the previous term to help expand it. Try writing out a few more terms and see what patterns you can find.
Jameson already did the groundwork by providing that excellent link. I did some work and here is what I got:What is the formula and how to derive the formula of sina+sin2a+sin3a...........?
sin(α) + sin (α + β) + sin (α + 2β) + ... to n terms
= [sin[α + (n - 1)β/2] . sin(nβ/2)]/sin(β/2)
This was a general case for which I derived a formula. Things become simpler for the formula which you require as we just need to put β = α and plug into the above formula.