1general soln of abv eqn is x=e^{i\theta } (-θ is also a soln,it will also work)
substitute in expression u'll get (cosn\theta + isin(n\theta) )(cosn\theta - isin(n\theta) ) = 1
if x^{2}-2xcos\theta +1=0 find the value ofx^{n}(2cosn\theta -x^{n}) where n is a natural number
-
UP 0 DOWN 0 0 4
4 Answers
1
21Jus put n=1
no need for ne other sub.
U directly get it frm the given eqtn
[4]
