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Published byRonald Barnett Modified over 9 years ago
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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
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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
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