In the trigonometric identity $\cos(\frac{\pi}{2} -\theta)$, why are we reflecting the graph in vertical axis. I was wondering why do we need to reflect the graph in vertical axis in the trigonometric identity: 

$\cos(\frac{\pi}{2} -\theta) = \sin(\theta)$.

It seems that if we only translate the graph of $\cos(\theta)$ by $\frac{\pi}{2}$ it would take the same values as the graph of $\sin(\theta)$, so why do we reflect the graph by changing sign of $\theta$ to negative.
 A: Hint:$$\cos(-x)=\cos(x)$$Cosine is even and that is one of the properties of an even function.
Note that $$\cos(\frac{\pi}2+\theta)\ne\sin(\theta)$$
A: It isn't that $\cos(\theta - \pi/2) = \sin(\theta)$ is more correct or less correct than $\cos(\pi/2 - \theta) = \sin(\theta)$: both are true.  However, the second one also expresses the familiar concept of "complementary angles": in any right triangle, the sine of one of the two (non-right) angles is always equal to the cosine of the other angle.  This follows from elementary opposite/adjacent/hypotenuse considerations, and thus makes it a more natural way to remember the identity.
Because of this naturality, the principle also generalizes much more readily than a simple shift would:
$$\cos(\pi/2 - \theta) = \sin(\theta)$$
$$\sin(\pi/2 - \theta) = \cos(\theta)$$
$$\cot(\pi/2 - \theta) = \tan(\theta)$$
$$\tan(\pi/2 - \theta) = \cot(\theta)$$
$$\csc(\pi/2 - \theta) = \sec(\theta)$$
$$\sec(\pi/2 - \theta) = \csc(\theta)$$
If you tried to express these in terms of $\theta \pm \pi/2$ you'd have to fiddle with the signs in each case to get it right.
