已知cos(π/2+x)=sin(x-π/2) 求sin^3(π-x)+cos(x+π)/5cos(5π/2-x)+3sin(7π/2-x)
问题描述：
已知cos(π/2+x)=sin(x-π/2) 求sin^3(π-x)+cos(x+π)/5cos(5π/2-x)+3sin(7π/2-x)
答
由cos(π/2+x)=sin(x-π/2) 得
-sinx=-cosx
sinx=cosx
sin^3(π-x)+cos(x+π)/5cos(5π/2-x)+3sin(7π/2-x)
=sin^3x--cosx/5sinx+3cosx
=(sin^2x--1)/8
又因sin^2x+cos^2x=1
sin^2=1/2
(sin^2x--1)/8=-3/32
即原式=-3/32
答
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)+3si...