The Taylor Expansion of $f(x)=\sin x$ with a Lagrange remainder is:
$\sin x = x-{x{3}\over 3!}+{x^{5}\over5!}+\cdots+{(-1)^{m-1}x^{2m-1}\over(2m-1)!}+{(-1)^{m}x^{2m+1}cos \theta x\over(2m+1)!}, 0<\theta<1, -\infty
which actually contains $2m$ terms and one $R(x)$ since $f^{(2k)}(x)=sin^{(2k)} x=0$:
$\sin x = x+0-{x{3}\over 3!}+0+{x^{5}\over5!}+0+\cdots+{(-1)^{m-1}x^{2m-1}\over(2m-1)!}+0+{(-1)^{m}x^{2m+1}cos \theta x\over(2m+1)!}, 0<\theta<1, -\infty
That's what I find in most maths books.
My question is:
Must I always regard the Taylor Expansion of $\sin x$ as containing $2m$ terms and one $R(x)$ ?
If the expansion contains only $2m-1$ terms and the $R(x)$, then $R(x)$ is the $2m$th term. So how can I write the $R(x)$ in Lagrange form (Obviously $R(x)$ is not equal zero)? Or I shouldn't do that ?
Any help will be great appreciated.