Question

Use the Maclaurin series for \(cosx\) to compute \(cos{5^o}\) correct

to five decimal places.

Short answer

\(\cos {5^ \circ } \approx 0.99619\)

Step-by-step solution

Step 1: Conversion to Radians

Recall that:

\(\begin{aligned}{l}{180^ \circ } = \pi rad\\\frac{5}{{180}} \times {180^ \circ } = \frac{5}{{180}} \times \pi rad\\{5^ \circ } = \frac{\pi }{{36}}radians\end{aligned}\)

Step 2: Maclaurin series

We know that:

\(\begin{aligned}{l}\cos x = \sum n = {0^\infty }{( - 1)^n}\frac{{{x^{2n}}}}{{(2n)!}}\\x = \frac{\pi }{{36}}\\\cos \left( {{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/ {\vphantom {\pi {36}}}\right.\\} \!\lower0.7ex\hbox{${36}$}}} \right) = \sum {\left( {{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/ {\vphantom {\pi {36}}}\right.\\} \!\lower0.7ex\hbox{${36}$}}} \right)} = {0^\infty }{( - 1)^n}\frac{{{{\left( {{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/ {\vphantom {\pi {36}}}\right.\\} \!\lower0.7ex\hbox{${36}$}}} \right)}^{2n}}}}{{(2n)!}}\end{aligned}\)

Step 3: The Alternating Series

The alternating series estimation theorem states that:

If we use the first n terms to estimate the sum of an infinite alternating series that is convergent, then the magnitude of error in that estimation is less than or equal to the magnitude of the (n+1)st term.

\(\left| {{R_n}} \right| \le \left| {{a_{n + 1}}} \right|\)

It is an alternating series. So we need to find the term less than \(0.00001\).

\(n\)	\({\raise0.7ex\hbox{${{{\left( {{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/ {\vphantom {\pi {36}}}\right.\\} \!\lower0.7ex\hbox{${36}$}}} \right)}^{2n}}}$} \!\mathord{\left/ {\vphantom {{{{\left( {{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/ {\vphantom {\pi {36}}}\right.\\} \!\lower0.7ex\hbox{${36}$}}} \right)}^{2n}}} {(2n)!}}}\right.\\} \!\lower0.7ex\hbox{${(2n)!}$}}\)
\(0\)	\(1\)
\(1\)	\(0.003808\)
\(2\)	\(2.4 \times {10^{ - 6}}\)

The term corresponding to \(n = 2\) is less than \(0.00001\), so we only need to add all the terms before it.

Use the magnitude of terms for \(n = 0\;\& \;n = 1\) from the table above

\(\cos {5^ \circ } \approx 1 - 0.003808 \approx 0.99619\)