arccos(sinx)与arcsin(cosx)如何比较大小

设y1=arccos(sinx), y2=arcsin(cosx)
则cosy1=sinx (0≤y1≤π/2)
siny2=cosx(-π/2≤y2≤π/2) ;
∴y1=π/2-x+2kπ或x-π/2+2kπ
y2=π/2-x+2kπ或x+π/2+2kπ;
当0≤y1,y2≤π/2时，即
0≤π/2-x+2kπ≤π/2或0≤x-π/2+2kπ≤π/2 ①
0≤π/2-x+2kπ≤π/2或0≤x+π/2+2kπ≤π/2 ②
y1=y2;
解①②得x∈[2kπ,π/2+2kπ]
x ∈(π/2+2kπ,2π+2kπ)时，y1∈(π/2+2kπ,π+2kπ)
y2∈(-π/2+2kπ,2kπ)
∴y2综上所述, x∈[2kπ,π/2+2kπ]时,y1=y2
x ∈(π/2+2kπ,2π+2kπ)时,y2