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18:30 90 t) f (c) = — + e (an cos na + bn sin nc) o, …...f (c) = — + e (an cos na + bn sin nc)...

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  • 18:30 90 T) f (c) = — + E (an cos na + bn sin nc) O, …...f (c) = — + E (an cos na + bn sin nc) O, bn(n = 1, 2, f@) cos nadc f (c) sin ncdc b, (X.) an, (1) l) an (n = 2 c e e
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Page 1: 18:30 90 T) f (c) = — + E (an cos na + bn sin nc) O, …...f (c) = — + E (an cos na + bn sin nc) O, bn(n = 1, 2, f@) cos nadc f (c) sin ncdc b, (X.) an, (1) l) an (n = 2 c e e
Page 2: 18:30 90 T) f (c) = — + E (an cos na + bn sin nc) O, …...f (c) = — + E (an cos na + bn sin nc) O, bn(n = 1, 2, f@) cos nadc f (c) sin ncdc b, (X.) an, (1) l) an (n = 2 c e e

Ex110 p 96 f(x)=cos(2x)-2cos(x) · 2016-01-30 · f(x+2𝝅)= cos(2x+4𝝅) -2 cos( x +2𝝅) = cos(2x) -2cos(x)=f(x) donc f est périodique de période 2𝝅 on pourra effectuer

izfrykse f=kdks.kferh; iQyu...izfrykse f=kdks.kferh; iQyu 23 mnkgj.k 10 tan (cos–1x) dk eku Kkr dhft, vkSj fiQj tan cos–1 8 17 ifjdfyr dhft,A gy –1eku yhft, cos x = θ, rc cos

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Algu m a tem um transformador para emprestar?F = B · A · cos q Þ F = 0,5 · 0,02 · cos 45” Þ F = 0,01 · 0,71 F = 0,0071 Wb. 46 AULA 2. Na Figura 4, suponha que uma bobina formada

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