What is the limit of this trig function? How do I find $$\lim_{x \to \pi/4}{\frac{\cos x-\frac{1}{\sqrt2}}{x-\frac\pi4}}$$? 
I've tried setting the denominator equal to $h$, then replacing $x$ in terms of $h$, but I still don't know how to manipulate it to look like $\frac{\cos h-1}h$.

As other have said, that is the definition of derivative of $\cos x$ at $x = \pi/4$.
To see why is that, set $h = x - \pi/4$, then you get 
$$\lim_{h \to 0} \frac{\cos(\pi/4 + h) - \cos(\pi/4)}h$$
So if you know that the derivative of $\cos x $ is $-\sin x$, you are done.
On may argue though that this is like cheating; so how to prove the limit without using the definition of derivative then?
rewrite the expression as $$\frac{\cos h \cos \pi/4 - \sin h \sin \pi/4 - \cos \pi/4}h = \frac{\frac{1}{\sqrt{2}} (\cos h - \sin h - 1)}h = \frac{1}{\sqrt 2} \frac{\cos h - 1}{h} - \frac{1}{ \sqrt 2} \frac{\sin h}h$$ 
And you know that $\lim_{h \to 0}\frac{\sin h}h = 1$, and $\lim_{h \to 0} \frac{\cos h - 1}{h}  = 0$, so your limit is $-\frac{1}{ \sqrt 2}$
