#equation353#
Next if we evaluate both sides at x=0, all terms except the first on the right hand side
vanish and we are left with
#equation355#
But we know that cos(0) = 1. Consequently, we have found that #tex2html_wrap_inline442#. Incidentally, we
also know that #tex2html_wrap_inline444# must vanish since sin(0) = 0 and all terms in the right hand side vanish
at x=0 except for #tex2html_wrap_inline450#. Thus for the two sides to be equal we thus need #tex2html_wrap_inline452# to be zero.
What about the other terms in the infinite series representation of sin(x)? These can be determined
by following a similar line. Namely, if we take two derivatives of both sides we find:
#equation357#
Again, if we evaluate both sides at x=0, we find that #tex2html_wrap_inline458#. Then, taking three derivatives
of both sides gives:
#equation359#
Then, evaluating both sides at x=0 yields #tex2html_wrap_inline462#.
Thus far we have found the series representation of sin(x) to be
#equation361#
Now we can see the strategy. To find the kth coefficient in the series expansion for
sin(x) near x=0, we take k derivatives of the function and evaluate both sides at
the <#364#>base point<#364#> x=0. Clearly, the result will be
#equation365#
In general, we might not want to find the series representation of a function
about the base point x=0 but rather a general point, say #tex2html_wrap_inline478#. What is the
form of the series representation then? The answer is
#equation372#
This is the Taylor series for f(x) about #tex2html_wrap_inline482#.