Question

Use the information from Exercise 5 to estimate cos 80^o
correct to five decimal places.

Short answer

The value up to five decimal places is 0.17365.

Step-by-step solution

Definition of decimal

A decimal number is defined as a number with a decimal point separating the full number and fractional parts. A decimal point is the dot in a decimal number. The digits after the decimal point represent a value less than one.

Calculate Taylor inequality

Let us calculate f(x),

\[\begin{array}{c}{\rm{f(x) = cosx,}}\\{\rm{a = }}\frac{{\rm{\pi }}}{{\rm{2}}}\\f(x) \approx {T_3}(x) \approx - \left( {x - \frac{\pi }{2}} \right) + \frac{1}{6}{\left( {x - \frac{\pi }{2}} \right)^3}\end{array}\]

\[\frac{\pi }{2} \approx 1.57079632679\]

So, let us check the Taylor inequality for interval

\[\frac{{4\pi }}{9} \le x \le \frac{\pi }{2}\]

Calculate the radian pattern values

Now, calculate the radians,

\[\begin{array}{c}f(x) = \cos x\\f\left( {\frac{\pi }{2}} \right) = 0\\{f^\prime }(x) = - \sin x\\{f^\prime }\left( {\frac{\pi }{2}} \right) = - 1\\{f^{\prime \prime }}(x) = - \cos x\\{f^{\prime \prime }}\left( {\frac{\pi }{2}} \right) = 0\\{f^{\prime \prime \prime }}(x) = \sin x\\{f^{\prime \prime \prime }}\left( {\frac{\pi }{2}} \right) = 1\end{array}\]

\[\]
This pattern repeats, so

\[{f^{(n)}} = \pm \cos x\]or \[ \pm \sin x\]

The Taylor term for the 4 th derivative is 0 so to check the accuracy we look at the next non-zero term, which uses the 5 th derivative

\[{f^{(5)}}(x) = - \sin x\]

The maximum of \[| - \sin x|\] over the interval \[\frac{{4\pi }}{9} \le x \le \frac{\pi }{2}\] is 1 , so we use that as M

\[\begin{array}{*{20}{r}}{\left| {{R_n}(x)} \right| \le \frac{M}{{(n + 1)!}}|x - a{|^{n + 1}}}\\{\left| {{R_3}(x)} \right| \le \frac{1}{{5!}}{{\left| {\frac{{4\pi }}{9} - \frac{\pi }{2}} \right|}^5} \le 0.00000135}\end{array}\]

The remainder is \[{\rm{ < 0}}{\rm{.00001}}\] so this tells us we have 5 decimal place accuracy with just \[{{\rm{T}}_3}\] and don't need any more terms.

Estimate the decimal

Use 80^oin radians in the equation

\[{T_3}\left( {\frac{{4\pi }}{9}} \right) \approx - \left( {\frac{{4\pi }}{9} - \frac{\pi }{2}} \right) + \frac{1}{6}{\left( {\frac{{4\pi }}{9} - \frac{\pi }{2}} \right)^3} \approx 0.17365\]

Therefore, the value up to five decimal places is 0.17365.