標題:
MQ46 --- Series
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發問:
MQ46 --- SeriesDifficulty: 35%Evaluate ∏(k = 0 to n)cos2?x.
最佳解答:
cosx cos2x cos4x ... cos(2^n x) =sinx cosx cos2x cos4x ... cos(2^n x) / sinx =(1/2) sin2x cos2x cos4x ... cos(2^n x) / sinx =(1/2)^2 sin4x cos4x ... cos(2^n x) / sinx =(1/2)^3 sin8x cos8x... cos(2^n x) / sinx = ... =(1/2)^(n+1) sin[2^(n+1) x] / sinx
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