The previous answer contains mistakes. Here is the correct derivation.

First of all, the minus sign in front of a function #f(x)=-sin(x)#, when taking a derivative, would change the sign of a derivative of a function #f(x)=sin(x)# to an opposite. This is an easy theorem in the theory of limits: limit of a constant multiplied by a variable equals to this constant multiplied by a limit of a variable. So, let's find the derivative of #f(x)=sin(x)# and then multiply it by #-1#.

We have to start from the following statement about the limit of trigonometric function #f(x)=sin(x)# as its argument tends to zero:

#lim_(h->0)sin(h)/h=1#

Proof of this is purely geometrical and is based on a definition of a function #sin(x)#. There are many Web resources that contain a proof of this statement, like The Math Page.

Using this, we can calculate a derivative of #f(x)=sin(x)#:

#f'(x)=lim_(h->0) (sin(x+h)-sin(x))/h#

Using representation of a difference of #sin# functions as a product of #sin# and #cos# (see Unizor , *Trigonometry - Trig Sum of Angles - Problems 4*) ,

#f'(x)=lim_(h->0) (2*sin(h/2)cos(x+h/2))/h#

#f'(x)=lim_(h->0) sin(h/2)/(h/2)*lim_(h->0)cos(x+h/2)#

#f'(x)=1*cos(x)=cos(x)#

Therefore, derivative of #f(x)=-sin(x)# is #f'(x)=-cos(x)#.