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).
