Q. 74 - 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.

$\sin 1°$

Step-by-Step Solution

Answer

The approximate the value of $\sin 1°$up to three decimal places is $0.0175$

Also, to approximate the quantity $\sin 1^{°}$to within ${10}^{-6}$ of its value, the number of terms required is $2$

step 1 : Given information

Consider the function $\sin 1°$

step 2 : calculation

Let us first convert the degree into radian measure.

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

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

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

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

So, to find the Maclaurin series for the function $\sin \frac{\pi }{180}$, put $x=\frac{\pi }{180}$

Therefore, 

$f\left(x=\frac{\pi }{180}\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{\left(\frac{\pi }{180}\right)}^{2k+1}$

That is

$\sin \frac{\pi }{180}=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{\left(\frac{\pi }{180}\right)}^{2k+1}$

Now, to approximate the value of $\sin \frac{\pi }{180}$up to three decimal places, let us first write its corresponding Maclaurin series in expanded form.

So,

$\sin \frac{\pi }{180}=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{\left(\frac{\pi }{180}\right)}^{2k+1}$

$=\frac{\pi }{180}-\frac{1}{3!}{\left(\frac{\pi }{180}\right)}^3+\frac{1}{5!}{\left(\frac{\pi }{180}\right)}^5-\frac{1}{7!}{\left(\frac{\pi }{180}\right)}^7+\frac{1}{9!}{\left(\frac{\pi }{180}\right)}^9$

$=0.01749-0.000000892$

Therefore, the approximate the value of $\sin 1°$ up to three decimal places is $0.0175$

Also, to approximate the quantity $\sin 1°$ to within ${10}^{-6}$ of its value, the number of terms required is $2$