If f(x) is a real-valued function defined as f(x)=ln⁡(1−sin⁡x), then the graph of f(x) is

fπ2−x=ln⁡(1−cos⁡x) and fπ2+x=ln⁡(1−cos⁡x) Thus, fπ2+x=fπ2−x Thus, f(x) is symmetrical about line x=π2.