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Sonam Ghatode Grade: 11
        


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

Answers : (1)

Badiuddin askIITians.ismu Expert
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


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Badiuddin


 



7 years ago
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