# cos A + cos (A+B) + cos (A+2B)+…+ cos [A+(n-1)B] = cos [A+(n-1)B/2] sin nB/2 /sin B/2

Grade:11

## 1 Answers

Badiuddin askIITians.ismu Expert
148 Points
14 years ago

Dear Sonam

let

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

or S = Real part of [eiA + ei(A+B) + ei(A+2B)+…+ ei(A+(n-1)B)]

S = Re [eiA{1+ei(B) + ei(2B)+…+ ei((n-1)B)}]

= Re [eiA{1+ei(B) + {ei(B)}2+{ei(B)}3…+ {ei(B)}n-1) }]

=Re [eiA{{ei(B)}n-1}/{ei(B)-1}]

=Re [eiA{cosnB + isin nB-1}/{cosB + isinB -1}]

= Re [eiA{1-2sin2nB/2 + i2sin nB/2 cosnB/2  -1}/{1-2sin2B/2 + i2sinB/2 cosB/2  -1}]

=Re [eiA{-2sin2nB/2 + i2sin nB/2 cosnB/2 }/{-2sin2B/2 + i2sinB/2 cosB/2  }]

=Re [eiA2isin nB/2{ cosnB/2 +isin nB/2 }/2isinB/2{cosB/2 +isin B/2  }]

=Re [sin nB/2 (cosA+isinA){ cosnB/2 +isin nB/2 }/sinB/2{cosB/2 +isin B/2  }]

=sin (nB/2)  /sin(B/2) Re [ cos(A+(n-1)B/2) +isin(A+(n-1)B/2)

compair real part

S =cos [A+(n-1)B/2] sin nB/2 /sin B/2

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Badiuddin

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