invers fungsi dari cos (x-phi/2)

[tex]y=\cos{(x-\frac{\pi}{2})} \\ y=\sin(x) \\ x=\arcsin(y) \\ f^{-1}(x)=\arcsin(x)[/tex]

[tex]f(x)=\sqrt{x-1},g(x)=1-x^2 \\ (f+g)(x)=f(x)+g(x)=\sqrt{x-1}+(1-x^2) \\ (f-g)(x)=f(x)-g(x)=\sqrt{x-1}-(1-x^2) \\ (f.g)(x)=f(x).g(x)=(1-x^2)\sqrt{x-1} \\ (f/g)(x)=f(x)/g(x)=\frac{\sqrt{x-1}}{1-x^2}=-\frac{\sqrt{x-1}}{x^2-1} \\ -\frac{\sqrt{x-1}}{(x+1)(x-1)}=-\frac{1}{(x+1)\sqrt{x-1}}[/tex]