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Question Number 163798 by kdaramaths last updated on 10/Jan/22
byasimplemethodcalculate∑k=0nCOS(kx)
Answered by Ar Brandon last updated on 10/Jan/22
Q162872
Answered by Kamel last updated on 10/Jan/22
byasimplemethodcalculateSn=∑k=0ncos(kx),Tn=∑nk=0sin(kx)(cos(x)−1)Sn+sin(x)Tn=∑nk=0cos((k−1)x)=cos(x)−cos(nx)−sin(x)Sn+(cos(x)−1)Tn=∑nk=0sin((k−1)x)=−sin(x)−sin(nx)∴{2sin2(x2)Sn−2sin(x2)cos(x2)Tn=cos(nx)−cos(x)...(1)2sin(x2)cos(x2)Sn+2sin2(x2)Tn=(sin(x)+sin(nx))...(2)(1).sin(x2)+(2)cos(x2)⇒Sn=sin(x2)cos(nx)−sin(x2)cos(x)+cos(x2)sin(x)+cos(x2)sin(nx)2sin(x2)∴Sn=sin((n+12)x)+sin(x2)2sin(x2)=sin(n+12x)cos(nx2)sin(x2)
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