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I have to do the following approximation using the Reminder Estimation Theorem, and my question is my answer right?:

$$\cos(.1) \approx \mathbf{some \ value}$$

(I have a calculator, and google, I know such value, and the units are in radians.)

Using the MacLaurin Series:

$$\cos(x)=\sum_\limits{n=0}^\infty\dfrac{(-1)^n x^{2n}}{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}…\frac{(-1)^n x^{2n}}{2n}$$

I then decided to substitute in the $.1$ for the $x$, and got the following.

$$\cos(.1)=\sum_\limits{n=0}^\infty\dfrac{(-1)^n .1^{2n}}{2n}=1-\frac{.1^2}{2!}+\frac{.1^4}{4!}…\frac{(-1)^n .1^{2n}}{2n}$$

The textbook that I am using gives the Reminder Estimation Theorem in the following format:

$$\vert R_n(x)\vert \le \frac{M}{(n+1)!}\vert x-x_o\vert^{n+1}$$

I then got it to the following method:

Method of Answering Question

\begin{align} 0\le\vert R_n(x)\vert&\le\frac{(.1)^{n+1}}{(n+1)!}\le.000005 \end{align} Using the first two terms I then processed the answer as the following: $$1-\frac{.1^2}{2!}=.995$$ Because the third term gave me the following value: $\frac{.1^4}{4!}\approx. 4.1666666...*10^{-6}$

Note these results are from a TI-83 Plus, and its in float mode.

Is my answer right?

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You are (partially) right. Either use the remainder, where $M$ is a bound on a derivative of $\cos x$ (thus 1), or see that it is a series of alternating sign terms, and the error is less than the first omited term.

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