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simplify-S-n-k-0-n-sin-k-x-cos-kx-




Question Number 60682 by maxmathsup by imad last updated on 24/May/19
simplify    S_n =Σ_(k=0) ^n   sin^k (x)cos(kx)
simplifySn=k=0nsink(x)cos(kx)
Commented by mathsolverby Abdo last updated on 29/May/19
we have S_n =Re(Σ_(k=0) ^n   sin^k x e^(ikx) )  =Re(Σ_(k=0) ^n (e^(ix) sinx)^k )  Σ_(k=0) ^n (e^(ix) sinx)^k  =((1−(e^(ix) sinx)^(n+1) )/(1−e^(ix) sinx))  =((1−sin^(n+1) x e^(i(n+1)x) )/(1−sinx(cosx +isinx)))  =((1−e^(i(n+1)x)  sin^(n+1) x)/(1−sinxcosx−isin^2 x))  =(((1−sinxcosx +isin^2 x)(1−sin^(n+1) x(cos(n+1)x+isin(n+1)x))/((1−sinxcosx)^2  +sin^4 x))  =...we get Sn after extracting Re(of this  quantity...)
wehaveSn=Re(k=0nsinkxeikx)=Re(k=0n(eixsinx)k)k=0n(eixsinx)k=1(eixsinx)n+11eixsinx=1sinn+1xei(n+1)x1sinx(cosx+isinx)=1ei(n+1)xsinn+1x1sinxcosxisin2x=(1sinxcosx+isin2x)(1sinn+1x(cos(n+1)x+isin(n+1)x)(1sinxcosx)2+sin4x=wegetSnafterextractingRe(ofthisquantity)

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