33It is the imaginary part of eia+ei2a...........
how do we solve the summation of series problems
liek sina + sin 2a + sin 3a ... + sin na
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7 Answers
1sin a + sin(a+b) + sin (a+2b) +... sin(a+(n-1)b)
=let q.
then 2qsin(b/2)=cos(a-b/2)- cos(a+b/2)+cos(a+b/2)-...cos(a+(n-3/2)b) -cos(a+(n-1/2)b).
=2sin(a+((n-1)/2)b)* sin (nb/2).
here b=a, so ans..
2sin(a+nb/2)sin((n+1)b/2).
62dude this is good..
i like the complex power method.. it is easier and less dirtly.. actually not dirty at all.
Someone willing to post that solution?
1this is nothing but the imaginary part of
(cosa+isina)+(cosa+isina)2+....(cosa+isina)n
hence this is a geometric progression
and we get it as
(isina + cosa)((isina+cosa)n-1)/(isina+cosa-1)
=
(sina + cosa)(isinna-1+icosna)/(isina-1+cosa)
simplifying we get the imaginary term as =
2sin(a(n+1)/2)sin(na/2)