已知cos(π/2+x)=sin(x-π/2) 求sin^3(π-x)+cos(x+π)/5cos(5π/2-x)+3s

因为cos(π/2+x)=-sinx,
sin(x-π/2)=sin[π-(x-π/2)]=sin(π/2-x)=cosx,
由cos(π/2+x)=sin(x-π/2),得：
-sinx=cosx.
所以[sin^3(π-x)+cos(x+π)]/[5cos(5π/2-x)+3sin(7π/2-x)]
=[(sinx)^3+cosx]/[5cos(π/2-x)+3sin(-π/2-x)]
=[(-cosx)^3+cosx]/[5sinx-3cosx]
=-cosx[1-(cosx)^2]/[-8cosx]
=1/8*[1-(cosx)^2].
又-sinx=cosx,
(sinx)^2+(cosx)^2=1,
所以 2(cosx)^2=1,
(cosx)^2=1/2,
所以原式=1/8*[1-1/2]=1/16.