Question 188287
Note: {{{cos(x+h)=cos(x)cos(h)-sin(x)sin(h)}}} this is one of the many trig identities


Since {{{f(x) = cos(x)}}}, this means that {{{f(x+h)=cos(x+h)=cos(x)cos(h)-sin(x)sin(h)}}} or simply {{{f(x+h)=cos(x)cos(h)-sin(x)sin(h)}}}



{{{(f(x+h)-f(x))/h}}} Start with the given expression



{{{(( cos(x)cos(h)-sin(x)sin(h) )- cos(x))/h}}} Plug in {{{f(x) = cos(x)}}} and {{{f(x+h)=cos(x)cos(h)-sin(x)sin(h)}}}



{{{(cos(x)cos(h) - cos(x) -sin(x)sin(h) )/h }}} Remove the parenthesis and rearrange the terms (I just moved the last cosine term right after the first term)



{{{(cos(x) (cos(h) - 1) -sin(x)sin(h))/h}}} Factor the term {{{cos(x)}}} from the first two terms.



{{{(cos(x) (cos(h) - 1))/h -(sin(x)sin(h))/h}}} Break up the fraction



So {{{(f(x+h)-f(x))/h=(cos(x) (cos(h) - 1))/h -(sin(x)sin(h))/h}}} where {{{f(x)=cos(x)}}}



Note: if you are in Calculus, then this would show that the derivative of {{{cos(x)}}} is {{{-sin(x)}}} (with a few more steps).