Calculus: Early Transcendentals, Chapter 3, 3.3, Section 3.3, Problem 20

By using the definition of the derivative,
lim_(h -> 0) [ cos(x + h) - cos(x) ] / h
=lim_(h -> 0)[ cos(x)cos(h) - sin(x)sin(h) - cos(x) ] / h
=lim_(h -> 0)[ cos(x) [ cos(h) - 1 ] - sin(x)sin(h) ] / h
=lim_(h -> 0)[ cos(x) [cos(h) - 1]/h ] - lim_(h -> 0) (sin(x)sin(h)/h)
=cos(x)lim_(h -> 0)[ [cos(h) - 1]/h ] - sin(x)lim_(h -> 0) (sin(h)/h)
as
lim_(h -> 0)[ [cos(h) - 1]/h ] =0 and
lim_(h -> 0) (sin(h)/h) = 1
= cos(x) (0) -sin(x) (1)
= -sin(x)