Stuck on trig simplification Stuck on trig simplification..
How do you reduce  $$\frac{\cos x-1}{x}$$ 
to $$\frac{-\sin x}{1}$$
?
What do I multiply by?
 A: If you are taking the limit $$\lim_{x\to 0} \frac{\cos(x)-1}{x},$$ which is of indeterminate form as $x \to 0$,  then it looks like in the subsequent step, L'Hopital was used to derive the equivalent limit:  $$\lim_{x\to 0} \frac{\cos(x)-1}{x}= \lim_{x\to 0} (-\sin x) = 0$$ i.e.  finding the derivative of both the numerator and denominator, and then evaluating the limit of the resultant function as $x\to 0$.
A: You use $x=x-0$ and $\cos(0)=1$ to see that your expression is the difference quotient of the cosine function
$\frac{\cos(x)-\cos(0)}{x-0}$
and that its limit for $x\to 0$ is the differential quotient or the derivative value at $x=0$.

Remark (expanded from comment): To use l'Hopital in this instance (supposing the original task did involve a limit of $x\to 0$) is circular reasoning. One has to determine the derivatives of numerator and denominator, but the derivative of the numerator at $x=0$ is exactly this limit. 
To have an easy-to-remember method, application of l'Hopital on limits of the form $\tfrac{f(x)}{x}$ is ok, but theoretically it is not sound. 

Another interpretation: There is really no limit intended in the question. Then a similar looking result is obtained from an application of the intermediate value theorem. In general
$$\frac{f(x)-f(a)}{x-a}=f'(\tilde x)$$
for $\tilde x$ in between $a$ and $x$, sometimes formalized as $\tilde x=(1-\theta)a+\theta x$ with $\theta \in (0,1)$.
In this instance, this gives 
$$\frac{\cos x-1}{x}=-\sin \tilde x,\quad \tilde x=\theta x,\quad\theta \in(0,1).$$
Exploration of the power series expressions on both sides gives $\tilde x=\frac{x}{2}+\frac{x^3}{12}+O(x^5)$ for sufficiently small values of $x$.
A: You can't reduce one to the other by any correct steps. This is because
$$\frac{\cos x -1}{x} \not\equiv -\sin x \, . $$
To see this, let $x=\pi$. We have:
$$\frac{\cos \pi - 1}{\pi} = -\frac{2}{\pi} \ \ \text{and} \ \ -\sin \pi = 0 \, . $$
I suspect that you are interested in the limit of $\frac{\cos x-1}{x}$ as $x$ tends towards zero. 
Notice that a limit is very different to a function. The two expressions might have the same limit, i.e. they tend towards the same value at the same point, but they need to be the same for all $x$. For example $x^2$ and $x^3$ both go to zero as $x$ goes to zero, but $2^2 \neq 2^3$.
To find the limit, take a look at these two videos on YouTube:


*

*Showing that $\frac{\sin x}{x} \to 1$ as $x \to 0$.

*Showing that $\frac{\cos x -1}{x} \to 0$ as $x \to 0$.
A: $\displaystyle\lim_{x\to0} \dfrac{\cos x-1}{x}=\lim_{x\to0} \dfrac{-2\sin^2 \frac{x}2}{x}=\lim_{x\to0} \dfrac{-\sin \frac{x}2}{\frac{x}2}\cdot \sin \frac{x}2=-1\cdot\lim_{x\to0} \sin\frac{x}2=0$
where I've used $\cos^2 x=1-2\sin^2 \dfrac{x}2$ and $\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1.$
