By drawing a right triangle it is obvious that $\cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)$. I'm trying to prove to myself that this is true for all values of $\theta$ by following the reasoning on this page where the sine graph is inverted vertically by plugging in $-\theta$ and then shifted to the right by $\frac{\pi}{2}$ to get the cosine graph. To shift a graph to the right by $x$, don't you have to subtract $x$ from the input value?
I'm getting $\cos \theta = \sin\left(-\theta - \frac{\pi}{2}\right)$ which is of course wrong and Wolfram shows me is the vertically-inverted cosine graph.
I know this is ridiculously simple and something to do with the $-\theta$ parameter to $\sin$ meaning that a right shift then needs the shift value added, but I can't see clearly why.