# Proving $\cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)$ for all $\theta$

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 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.

Alternatively, one can make use of

$$\sin (a-b)=\sin a \cos b - \sin b \cos a$$

Then $a=\pi/2$ and also $b=\theta$,

$$\sin (\pi/2-\theta)=\sin \pi/2 \cos \theta - \sin \theta \cos \pi/2$$

$$\sin (\pi/2-\theta)=1 \cdot \cos \theta - \sin \theta \cdot0$$

$$\sin (\pi/2-\theta)=\cos \theta $$

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