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

Dear Sonam

let

S =cos A + cos (A+B) + cos (A+2B)+…+ cos [A+(n-1)B]

or S = Imaginary part of [e^iA + e^i(A+B) + e^i(A+2B)+…+ e^i(A+(n-1)B)]

   S = Im[e^iA{1+e^i(B) + e^i(2B)+…+ e^i((n-1)B)}]

      = Im [e^iA{1+e^i(B) + {e^i(B)}²+{e^i(B)}³…+ {e^i(B)}^n-1) }]

       =Im [e^iA{{e^i(B)}ⁿ-1}/{e^i(B)-1}]

       =Im [e^iA{cosnB + isin nB-1}/{cosB + isinB -1}]

       = IM [e^iA{1-2sin²nB/2 + i2sin nB/2 cosnB/2  -1}/{1-2sin²B/2 + i2sinB/2 cosB/2  -1}]

       =Im [e^iA{-2sin²nB/2 + i2sin nB/2 cosnB/2 }/{-2sin²B/2 + i2sinB/2 cosB/2  }]

       =Im [e^iA2isin 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

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