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LHS = Im ( n C 1 e i q + n C 2 e 2i q + n C 3 e 3i q ……. n C n e ni q )

Summing Trigometric Series ( Learn as it has come up). Ex1 Show n C 1 sin q + n C 2 sin2 q + n C 3 sin3 q + ………..sin( n q ) = 2 n cos n (  q )sin(  n q ). LHS = Im ( n C 1 e i q + n C 2 e 2i q + n C 3 e 3i q ……. n C n e ni q ). Z = cos q +isin q = e i q. Subst z = e i q.

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LHS = Im ( n C 1 e i q + n C 2 e 2i q + n C 3 e 3i q ……. n C n e ni q )

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  1. Summing Trigometric Series (Learn as it has come up) Ex1 Show nC1sinq + nC2sin2q + nC3sin3q + ………..sin(nq) = 2ncosn(q)sin(nq) LHS = Im(nC1eiq + nC2e2iq + nC3e3iq…….nCneniq) Z = cosq+isinq = eiq Subst z = eiq LHS = Im(1 + nC1z + nC2z2 + nC3z3 +…….nCnzn– 1) Add 1 at the start and subtract 1 at the end = Im[(1 + z)n – 1] by the binomial theorem (1 + x)n = 1 + nC1x + nC2x2 + nC3x3 …..nCnxn = Im[(1 + eiq)n – 1] = Im[(1 + cosq + isinq)n – 1] = Im[(1+2cos2q – 1 + i2sinqcosq)n – 1] = Im [(2cosq)n(cosq + isinq)n – 1] = Im[(2ncosnq)(cosnq + isinnq) – 1] = 2ncosnqsinnqProven Using only the imaginary part cos2q = 2cos2q –1 sin2q = 2sinqcosq

  2. Show 1 + cosq+ cos2q +cos3q+…..cos(n–1)q = LHS = Re(1 + eiq + e2iq + e3iq+ e4iq+ .....e(n–1)iq) Subst z = eiq = Re(1 + z + z2 + z3 +….zn–1) This is a G.P = Re Using the formula for the sum of a G.P = Re = Re Rationalise the denominator

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