Q. 75 - Calculus

Question

Use appropriate Maclaurin series to express the quantities in Exercises 67–76 as alternating series. Then use Theorem 7.38 to approximate the value of the specified quantities to within 0.001 of their actual value. How many terms in each series would be needed to approximate the given quantity to within 10−6 of its value? In Exercises 73–76 be sure to convert to radian measure first. 

$\cos 5°$

Step-by-Step Solution

Answer

The approximate the value of  $\cos 5°$ up to three decimal places is 0.996

Also, to approximate the quantity $\cos 5^{\text{'}}$ to within ${10}^{-6}$ of its value, the number of terms required is 3

Step 1: Given information

Consider the function $\cos 5°$

Step 2: Calculation

Let us first convert the degree into radian measure.

Since $\mathrm{\pi }$ radians =$180°$, therefore $1°=\frac{\pi }{180}$ radian

So,

$5°=\frac{5\pi }{180}$

=$\frac{\pi }{36}$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 $\cos \frac{\pi }{36}$, put $x=\frac{\pi }{36}$

Therefore,

$f\left(x=\frac{\pi }{36}\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{\left(\frac{\pi }{36}\right)}^{2k}$

That is

$\cos \frac{\pi }{36}=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{\left(\frac{\pi }{36}\right)}^{2k}$

Now, to approximate the value of $\cos \frac{\pi }{36}$ up to three decimal places, let us first write its corresponding Maclaurin series in expanded form.

So,

$$\begin{gathered} \cos \frac{\pi }{36}=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{\left(\frac{\pi }{36}\right)}^{2k} \\ =1-\frac{1}{2!}{\left(\frac{\pi }{36}\right)}^2+\frac{1}{4!}{\left(\frac{\pi }{36}\right)}^4-\frac{1}{6!}{\left(\frac{\pi }{36}\right)}^6+\frac{1}{8!}{\left(\frac{\pi }{36}\right)}^8 \\ =1-0.00381-0.00000242 \end{gathered}$$

Therefore, the approximate the value of $\cos 5$ up to three decimal places is $0.996$

Also, to approximate the quantity $\cos 5$ to within ${10}^{-6}$ of its value, the number of terms required is 3