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```
sin A +sin (A+B) +sin(A+2B)+…+sin[A+(n-1)B] = sin[A+(n-1)B/2] sin nB/2 /sin B/2
```
7 years ago

147 Points
```										Dear Sonam
let
S =cos A + cos (A+B) + cos  (A+2B)+…+ cos [A+(n-1)B]
or S = Imaginary part  of [eiA + ei(A+B) + ei(A+2B)+…+ ei(A+(n-1)B)]
S = Im[eiA{1+ei(B) + ei(2B)+…+ ei((n-1)B)}]
= Im  [eiA{1+ei(B) + {ei(B)}2+{ei(B)}3…+ {ei(B)}n-1) }]
=Im [eiA{{ei(B)}n-1}/{ei(B)-1}]
=Im  [eiA{cosnB  + isin nB-1}/{cosB +  isinB -1}]
= IM  [eiA{1-2sin2nB/2  +  i2sin nB/2 cosnB/2  -1}/{1-2sin2B/2 +  i2sinB/2 cosB/2  -1}]
=Im  [eiA{-2sin2nB/2   +  i2sin nB/2 cosnB/2 }/{-2sin2B/2 +  i2sinB/2 cosB/2  }]
=Im  [eiA2isin nB/2{ cosnB/2 +isin nB/2 }/2isinB/2{cosB/2  +isin B/2  }]
=Im  [sin nB/2 (cosA+isinA){ cosnB/2 +isin nB/2 }/sinB/2{cosB/2   +isin B/2  }]
=sin  (nB/2)  /sin(B/2) Im [ cos(A+(n-1)B/2)  +isin(A+(n-1)B/2)

compair imaginary  part
S =sin[A+(n-1)B/2] sin nB/2 /sin B/2
Please feel  free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we  will get you  the answer and detailed  solution very  quickly.   We are all IITians and here to help you in your IIT JEE preparation.  All the best.   Regards, Askiitians Experts Badiuddin

```
7 years ago
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