Innequality prove question

transgalactic

prove that..
$$x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x\\$$
$$R_5=\frac{f^{5}(c)x^5}{5!}$$
i need to prove that the remainder is negative .
$$\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5$$
$$R_5=\frac{cos(c)x^5}{5!}$$

pandagoat

I'm going to guess that you will use the Maclaurin series for sin(x) and cos(x).

transgalactic

no?????/////

i need to show that the remainder is negative??

Mark44

Mentor
$$x-\frac{x^3}{6}+\frac{x^5}{120}$$
These are the first three terms of the Maclaurin series for sin(x). Do you know what the terms immediately following them look like? Do you know how successive terms in this Maclaurin series compare in absolute value?

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