Q. 76 - Calculus

Question

Use appropriate Maclaurin series to find the first four nonzero terms in the Maclaurin series for the product functions in Exercises 61-66. Also, give the interval of convergence for the series.

$\cos 10°$

Step-by-Step Solution

Answer

Therefore, the approximate the value of $\cos 10°$ up to three decimal places is$0.985$

Also, to approximate the quantity $\cos 10°$ to vithin ${10}^{-6}$of its value, the number of terms required is$4$

Step 1: Given information

Consider the function $\cos 10°$

Step 2: Calculation

Let us first convert the degree into radian measure. Since$\pi radians=180°,therefore1°=\frac{\pi }{180}radian$ 

So,

$=\frac{\mathrm{\pi }}{18}radian$

Also, let us consider$f\left(x\right)=\cos x$

The Maclaurin series for the function $f\left(x\right)=\cos x$ is

$f\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}$

So, to find the Maclaurin series for the function$\cos \frac{\pi }{18},putx=\frac{\pi }{18}$Therefore,

$f\left(x=\frac{\pi }{18}\right)=\sum _{i=0}^{\kappa }\frac{(-1{)}^k}{\left(2k\right)!}{\left(\frac{\pi }{18}\right)}^{2k}$

That is

$\cos \frac{\pi }{18}=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{\left(\frac{\pi }{18}\right)}^{2k}$

Now, to approximate the value of $\cos \frac{\pi }{18}$ up to three decimal places, let us first write its corresponding Maclaurin series in expanded form.

So,

$$\begin{gathered} \cos \frac{\pi }{18}=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{\left(\frac{\pi }{18}\right)}^{2k} \\ =1-\frac{1}{2!}{\left(\frac{\pi }{18}\right)}^2+\frac{1}{4!}{\left(\frac{\pi }{18}\right)}^4-\frac{1}{6!}{\left(\frac{\pi }{18}\right)}^6+\frac{1}{8!}{\left(\frac{\pi }{18}\right)}^8 \\ =1-0.01524-0.0000387 \end{gathered}$$

Therefore, the approximate the value of $\cos 10°$ up to three decimal places is $0.985$

Also, to approximate the quantity $\cos 10°$ to within ${10}^{-6}$ of its value, the number of terms required is $4$