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I have to use the derivative definition to determine the derivability of a function at a point. After trying Wolframalpha to help me out, it shows this step, but I don't understand how this:

$$\lim_{h\to 0}\frac{2\cos\left(\frac{\pi+h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}$$ turns into this: $$\lim_{h\to 0}\frac{\cos\left(h\right)-1}{h} $$

I don't recognize a trig identity here. Please explain the magic to me, and help me to solve the limit.

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  • $\begingroup$ $\cos (\frac \pi 2+\alpha)=\ldots$ $\endgroup$ – Canis Lupus Oct 28 '16 at 21:00
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HINT: Use the facts that $\cos\left(\frac{\pi + h}{2}\right) = - \sin\left(\frac h2\right)$ and $2 \sin^2\left(\frac{\alpha}{2}\right) = 1- \cos(\alpha)$

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Okay. First you have to see this.

$\cos\left(\frac{\pi + h}{2}\right) = - \sin\left(\frac h2\right)$

Then you can see that your trigonometry part of the given limit can rewrite as,

$-2\sin\left(\frac h 2\right)\sin\left(\frac{h}{2}\right) = -2 \sin^2\left(\frac{h}{2}\right) $

Now again you have to use this,

$ \cos(h) = 1-2 \sin^2\left(\frac{h}{2}\right) $

Then clearly you can see that what you got is equal to what we calculated here.

$\cos\left(h\right)-1$

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