2.4.4 So­lu­tion math­ops-d


The in­ver­sion, or par­ity, op­er­a­tor ${\mit\Pi}$ turns $f(x)$ into $f(-x)$. (It plays a part in the ques­tion to what ex­tent physics looks the same when seen in the mir­ror.) Show that ${\mit\Pi}$ leaves $\cos(x)$ un­changed, but turns $\sin(x)$ into $\vphantom{0}\raisebox{1.5pt}{$-$}$$\sin(x)$.


Ac­cord­ing to [1, p. 43], $\cos(-x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\cos(x)$, but $\sin(-x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$\sin(x)$. Com­pare also the graphs of the func­tions on the same page; ${\mit\Pi}$ flips the graph of a func­tion over around the $y$-​axis.