<s>[INST] Finding the derivative of sinus and cosinus. Trigonometric identities How can we see that $$\sin(x+h)-\sin(x)=2\sin\left(\frac h2\right)\cos\left(x+\frac h2\right)$$
How can we see that $$\cos(x+h)-\cos(x)=-2\sin\left(\frac h2\right)\sin\left(x+\frac h2\right)$$
Do these identities have a name?
 [/INST] You asked for a name: Looks like an application of the reverse of the Prosthaphaeresis identities, or sum-to-product identities.
$$
\sin a − \sin b = 2 \cos\frac{a + b}{2} \sin\frac{a − b}{2} \\
\cos a − \cos b = −2 \sin\frac{a + b}{2} \sin\frac{a − b}{2}
$$
using $a = x + h$ and $b = x$.
Another link here.
I usually use the complex definitions of $\sin$ and $\cos$ to proof, but you probably want an elementary one.
Note there are four identities in total.
$$
\sin a + \sin b = 2 \sin\frac{a + b}{2} \cos\frac{a − b}{2} \\
\cos a + \cos b = 2 \cos\frac{a + b}{2} \cos\frac{a − b}{2}
$$
which would yield
$$
\sin(x+h) + \sin(x) = 
2 \sin\left(x+\frac{h}{2}\right) \cos\left(\frac{h}{2}\right) \\
\cos(x+h) + \cos(x) = 
2 \cos\left(x+\frac{h}{2}\right) \cos\left(\frac{h}{2}\right)
$$
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