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?
 A: For any $a,b$, we have the well known "addition theorems"
$$ \sin(a+b) = \sin a \cos b + \sin b \cos a$$
and $$ \sin(a-b) = \sin a \cos b - \sin b \cos a $$
Subtracting these two equations, we get 
$$ \sin(a+b) - \sin(a-b) = 2\sin b \cos a $$
For the cosine, we have
$$ \cos(a+b) = \cos a \cos b - \sin a \sin b $$
and $$ \cos(a-b) = \cos a \cos b + \sin a \sin b $$
Subtraction again 
$$ \cos(a+b) - \cos(a-b) =  -2\sin a\sin b$$
Now let $b = \frac h2$, $a = x+\frac h2$.
A: 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
$$
\sin(x+h) + \sin(x) = 
2 \sin\left(x+\frac{h}{2}\right) \cos\left(\frac{h}{2}\right) \\
\cos(x+h) + \cos(x) = 
2 \cos\left(x+\frac{h}{2}\right) \cos\left(\frac{h}{2}\right)
$$
