Verifying $\cos\left(x-\frac{\pi}{2}\right)=\sin(x)$ without using other identities Say I forgot all the trig identities during an exam, including 
$$\cos(a-k)=\cos(a)\cos(k)-\sin(a)\sin(k)$$

Is there a way I can verify the following trig identity without using other identities, and without verifying by graphing?
  $$\cos\left(x-\frac{\pi}{2}\right)=\sin(x)$$

 A: If you go back to the first definition of $\sin$ and $\cos$ you learn, the ratio between sides in a right-angled triangle, it is quite obvious that the sine of one angle is equal to the cosine of the other.
A: This may not be the answer you're looking for (depending on your level), but you can derive most trig identities from the expressions
$$sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$$
and
$$cos(z) = \frac{e^{iz}+e^{-iz}}{2}$$
Then manipulating the above expressions using high-school algebra and Euler's formula. 
I wouldn't recommend it as a replacement for memorizing things for tests, but it's as close as a catch-all method for deriving trig identities as it gets. 
A: On the unit circle, $\theta \mapsto \pi/2-\theta$ corresponds to reflection across the $y=x$ line, hence interchanges $\cos$ and $\sin$,
$$\cos(\pi/2-\theta)=\sin(\theta).$$
The transformation $\theta \mapsto -\theta$ corresponds to reflection across the $y=0$ line, hence preserves $\cos \theta = x$,
$$\cos(-\theta) = \cos(\theta).$$
Now put those together.
