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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 $$f(\tfrac{\pi}{2}-x)=f(\tfrac{\pi}{2}+x).$$

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Thanks for all the help! So let's say you don't know the trig identities. Your triangle method is by far the best right? In my book they use the unit circle and draw it, but to me that's a bit vague – Trig Apr 5 '13 at 6:22
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

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