Question

函数y=cosx的图像在（0，π/2）上的部分

Answer 1

f(x)=arcsin(sinx)当x∈[-π/2,π/2]时，arcsin(sinx)=x当x∈(π/2,3π/2]时，π-x∈[-π/2,π/2)sinx=sin(π-x)∴arcsin(sinx)=π-x当x∈[2kπ-π/2,2kπ+π/2],k∈Zx-2kπ∈[-π/2,π/2],k∈Z∴arcsin(sinx)=x-2kπ,当x∈(2kπ+π/2,2kπ+3π/2],k∈Z(2k+1)π-x∈[-π/2,π/2)arcsin(sinx)=(2k+1)π-x即f(x)={x-2kπ,x∈[2kπ-π/2,2kπ+π/2],k∈Z{(2k+1)π-x,x∈(2kπ+π/2,2kπ+3π/2],k∈Z