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I know that \cos'(0) is $0$, but my work follows:

\begin{align}\cos'(0) &= \lim_{\Delta x\to 0}\frac{f(0+\Delta x) - f(0)}{\Delta x}\\ &= \frac{\cos(0+\Delta x) - \cos(0)}{\Delta x}\\ &= \frac{\cos(0) - \cos(0)}{0} = \frac{0}{0}\end{align}

Where have I gone wrong? What can I do to show that it equals 0?

(Note: No derivative rules are allowed in my calculus class yet, just difference quotients)

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    What esle did you do wrong? You "dropped" the lim in front, with no reason. Then you set $\Delta x$ equal to $0$ with no reason. But you will still need to know some non-trivial property of trig functions.2011-09-11

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You can't just substitute in 0 for the limit - that loses all important information and leaves you with an indeterminate form. You must be wittier - usually, with $\cos$ and $\sin$ and their derivatives (even with difference quotients), one ends up using the following inequalities.

$\cos A - \cos B = -2 \sin {\frac{1}{2} (A + B)} \cos {\frac{1}{2} (A + B)}$

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

Or you could use the difference of $\sin$ angles too, depending on how you tackle the proof. The key is to not get rid of the $\Delta x$ too early - evaluate the limit only when you come across a form that you know how to evaluate.

I think that gives you a next step, right?

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    @Thijs All sum-angle and sum-difference formulas, like the ones in the answer, can be derived geometrically (sometimes this is done using vectors in the plane). One can of course get the formulas for $\sin 2x$, $\cos 2x$ and so on as special cases. See [here](http://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Angle_sum_identities).2011-09-11
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Because it is easier to type, I will write $h$ instead of $\Delta x$. We want to find $\lim_{h \to 0} \frac{\cos h -1}{h}.$ Multiply "top" and "bottom" by $\cos h+1$. We want $\lim_{h \to 0} \frac{(\cos h -1)(\cos h+1)}{h(\cos h+1)}.$ On top we now have $-\cos^2 h +1$, that is, $-\sin^2 h$. So we want $\lim_{h \to 0} \frac{-\sin^2 h}{h(\cos h+1)}\qquad\text{that is,}\qquad \lim_{h \to 0} \left(\frac{\sin h}{h}\cdot \frac{-\sin h}{\cos h+1}\right).$

Finally, let $h \to 0$. We are allowed to use the fact that $\lim_{h\to 0}\frac{\sin h}{h}=1$. And it is clear that $\lim_{h\to 0}\frac{-\sin h}{\cos h +1}=0$. So our limit is $0$.

Comment: The idea used above is related to the process of "rationalizing the numerator," which you may have seen already, for example in finding the derivative of $\sqrt{x}$ at $x=3$ from the definition of the derivative.