1 / 3

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.

thad
Download Presentation

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 )

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related