Ex1Show n C 1 sin  + n C 2 sin2  + n C 3 sin3  + ………..sin(n  ) = 2 n cos n (   )sin(  n  ) LHS = Im( n C 1 e i  + n C 2 e 2i  + n C 3 e 3i  ……. n C n e ni  ) Subst z = e i  LHS = Im(1 + n C 1 z + n C 2 z 2 + n C 3 z 3 +……. n C n z n – 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 + n C 1 x + n C 2 x 2 + n C 3 x 3 ….. n C n x n = Im[(1 + e i  ) n – 1] = Im[(1 + cos  + isin  ) n – 1] = Im[(1+2cos 2   – 1 + i2sin   cos   ) n – 1] = Im [(2cos   ) n (cos   + isin   ) n – 1] = Im [(2 n cos n   )(cos  n  + isin  n  ) – 1] = 2 n cos n   sin  n  ProvenUsing only the imaginary part Summing Trigometric Series (Learn as it has come up) Z = cos  +isin  = e i  sin2  = 2sin  cos  cos2  = 2cos 2  –1

Show 1 + cos  + cos2  +cos3  +…..cos(n–1)  = LHS = Re(1 + e i   e 2i   e 3i   e 4i   e (n–1)i   Subst z = e i  = Re(1 + z + z 2 + z 3 +….z n–1 )This is a G.P = Re Using the formula for the sum of a G.P = Re Rationalise the denominator = Re