# How to show that $f(x) = \cos^2(x)\sin(x)$ is symmetric about the line $x=\frac{1}{2} \pi$?

I really have trouble with making any exercises regarding point symmetry and line symmetry. For example:

Show that $f(x) = \cos^2(x)\sin(x)$ is line symmetrical in the line $x=\dfrac{1}{2} \pi$.

So I need to show $f(a-p) = f(a+p)$, but I have no idea how to. Even when I see the answers, I don't know. I just see them rewriting cosines and sines in different forms.

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Note that $$\sin(\tfrac{\pi}{2}-x)=\cos(x)\qquad \text{ and }\qquad \cos(\tfrac{\pi}{2}-x)=\sin(x)$$ (see Wikipedia for example; you can also just draw a triangle with angles $x$, $\frac{\pi}{2}-x$, and $\frac{\pi}{2}$).
Using the above facts together with the facts $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$, we see that $$\sin(\tfrac{\pi}{2}+x)=\cos(x)\qquad\text{ and }\qquad \cos(\tfrac{\pi}{2}+x)=-\sin(x).$$ Thus, we have $$f(\tfrac{\pi}{2}-x)=\bigg[\cos(\tfrac{\pi}{2}-x)\bigg]^2\sin(\tfrac{\pi}{2}-x)=\sin^2(x)\cos(x)$$ and $$f(\tfrac{\pi}{2}+x)=\bigg[\cos(\tfrac{\pi}{2}+x)\bigg]^2\sin(\tfrac{\pi}{2}+x)=(-\sin^2(x))\cos(x)=\sin^2(x)\cos(x)$$ so that $$f(\tfrac{\pi}{2}-x)=f(\tfrac{\pi}{2}+x).$$
I suppose so; I certainly keep very few trig identities in memory, I just work it out when necessary. Sometimes drawing a triangle is the best way to go, but if you're comfortable with complex numbers and their relation with trigonometric functions, e.g. these identities and de Moivre's formula $$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx),$$ then that provides a very useful technique as well. – Zev Chonoles Apr 5 '13 at 6:28