Question 1152913
<pre>
Before you can prove that, you must first prove two other theorems.

{{{matrix(1,3,
matrix(2,2,
lim, (cos(h)-1)/h,
"h->0",""),
""="",
0)}}} 

and

{{{matrix(1,3,
matrix(2,2,
lim, (sin(h))/h,
"h->0",""),""="",1)}}}

I will assume you have already proved those.  If you haven't, post again,
asking for those proofs.

{{{matrix(1,3,"y'",""="",
matrix(2,2,
lim, (cos(x+h)-cos(x))/h,
"h->0",""))}}}

{{{matrix(1,3,"y'",""="",
matrix(2,2,
lim, (cos(x)cos(h)-sin(x)sin(h)-cos(x))/h,
"h->0",""))}}}

{{{matrix(1,3,"y'",""="",
matrix(2,2,
lim, (cos(x)cos(h)-cos(x)-sin(x)sin(h))/h,
"h->0",""))}}}

{{{matrix(1,3,"y'",""="",
matrix(2,2,
lim, (cos(x)(cos(h)-1^"")-sin(x)sin(h))/h,
"h->0",""))}}}

{{{matrix(1,3,"y'",""="",
matrix(2,2,
lim, (cos(x)((cos(h)-1)/h)^""-sin(x)((sin(h))/h)^""),
"h->0",""))}}}

Since we are assuming the two theorems at the top, the expression 

{{{(cos(h)-1)/h}}} 

approaches 0, and the expression

{{{sin(h)/h}}} 

approaches 1, so the expression approaches

{{{cos(x)*0-sin(x)*1}}}

which is

{{{-sin(x)}}}

Edwin</pre>