# Derivative of $\sin x$ for small $x$.

I have a question which requires me to prove from first principles that the derivative of $\sin x$ is $\cos x$. You may use $$\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$$

without proof. I managed to do this quite easily. But the second question stumped me:

When $x$ increases from $\pi$ to $\pi + \epsilon$, where $\epsilon$ is small, the increment in $\frac{\sin x}{x}$ is approximately equal to $p\epsilon$. Find $p$.

The given answer is $-\frac{1}{\pi}$, but I don't know how to get there. I know that for small $\theta$: $\sin\theta \approx \theta$, but I'm not sure how to apply that in this scenario.

• Hint: use the $\sin(a+b)$ with $a=\pi$ and $b=\epsilon$ – Martigan Oct 15 '14 at 15:12
• Doesn't that make $\sin\pi\cos\epsilon + \cos\pi\sin\epsilon$? Not sure how that helps :/ – hohner Oct 15 '14 at 15:15
• I think the answer below shows you how it helps... – Martigan Oct 15 '14 at 15:16

You can use $\dfrac{sin(\pi+\epsilon)}{\pi+\epsilon}=\dfrac{-sin(\epsilon)}{\pi+\epsilon}\approx\dfrac{-\epsilon}{\pi}$, because when $\epsilon\to 0$, you have $\sin(\epsilon)\approx\epsilon$ and $\pi+\epsilon\approx\pi$.
• Why does $\sin(\pi + \epsilon)$ equal $-\sin(\epsilon)$? – hohner Oct 15 '14 at 15:20
• for all $x$, $\sin(\pi+x)=-\sin(x)$ (look at the circle). – Denis Oct 15 '14 at 15:23
The idea is to use a symmetry of $\sin$ to transfer what you know about the infinitesimal behavior of $\sin x$ at $x = 0$ to $x = \pi$. In particular, $\sin(x + \pi) = -\sin(x)$, so $$\sin(\pi + \epsilon) = -\sin \epsilon = -\epsilon + O(\epsilon)^2.$$
Thus, $$\frac{\sin(\pi + \epsilon)}{\pi + \epsilon} = (-\epsilon + O(\epsilon^2))\left(\frac{1}{\pi} + O(\epsilon)\right) = -\frac{1}{\pi} \epsilon + O(\epsilon^2),$$ so as you suggest $$p = -\frac{1}{\pi}.$$