derivatives of sine and cosine

The of the derivatives of sine and cosine is a bit simpler by using the prosthaphaeresis formulas

$\sin \alpha -\sin \beta =\mathrm{\hspace{0.17em}2}\sin \left({\displaystyle \frac{\alpha -\beta }{2}}\right)\cos \left({\displaystyle \frac{\alpha +\beta }{2}}\right),$(1)

$\cos \alpha -\cos \beta =-2\sin \left({\displaystyle \frac{\alpha +\beta }{2}}\right)\sin \left({\displaystyle \frac{\alpha -\beta }{2}}\right).$(2)

Let $x,t$ be any real numbers such that  $t\ne x$.  Then we obtain

$$\frac{\sin x-\sin t}{x-t}=\frac{2\sin \left(\frac{x-t}{2}\right)\cos \left(\frac{x+t}{2}\right)}{x-t}=\frac{\sin \left(\frac{x-t}{2}\right)}{\left(\frac{x-t}{2}\right)}\cdot \cos \left(\frac{x+t}{2}\right)⟶\mathrm{\hspace{0.33em}1}\cdot \cos \left(\frac{x+x}{2}\right)=\cos x,$$

as  $t\to x$.  Here we used the known limit  $\underset{u\to 0}{lim}{\displaystyle \frac{\sin u}{u}}=1$  (see this entry (http://planetmath.org/LimitOfDisplaystyleFracsinXxAsXApproaches0)).

The derivative of cosine is calculated similarly:

$$\frac{\cos x-\cos t}{x-t}=\frac{-2\sin \left(\frac{x+t}{2}\right)\sin \left(\frac{x-t}{2}\right)}{x-t}=-1\cdot \frac{\sin \left(\frac{x-t}{2}\right)}{\left(\frac{x-t}{2}\right)}\cdot \sin \left(\frac{x+t}{2}\right)⟶-1\cdot 1\cdot \sin \left(\frac{x+x}{2}\right)=-\sin x,$$

as  $t\to x$.