Answer
${\sin ^{ - 1}}\dfrac{1}{2} \approx \dfrac{1}{2} + \dfrac{1}{2}\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^3}}}{3} + \dfrac{{1\cdot3}}{{2\cdot4}}\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^5}}}{5} + \dfrac{{1\cdot3\cdot5}}{{2\cdot4\cdot6}}\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^7}}}{7} + \dfrac{{1\cdot3\cdot5\cdot7}}{{2\cdot4\cdot6\cdot8}}\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^9}}}{9} \approx 0.523585$
Compare this result with the calculator value: ${\sin ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{6} \approx 0.523599$, the error is less than ${10^{ - 4}}$.
Work Step by Step
Recall from Exercise 43:
${\sin ^{ - 1}}x = x + \mathop \sum \limits_{n = 1}^\infty \dfrac{{1\cdot3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot4\cdot6\cdot\cdot\cdot\left( {2n} \right)}}\dfrac{{{x^{2n + 1}}}}{{2n + 1}}$
For the first five terms, we have the approximation:
${\sin ^{ - 1}}x \approx x + \dfrac{1}{2}\dfrac{{{x^3}}}{3} + \dfrac{{1\cdot3}}{{2\cdot4}}\dfrac{{{x^5}}}{5} + \dfrac{{1\cdot3\cdot5}}{{2\cdot4\cdot6}}\dfrac{{{x^7}}}{7} + \dfrac{{1\cdot3\cdot5\cdot7}}{{2\cdot4\cdot6\cdot8}}\dfrac{{{x^9}}}{9}$
Setting $x = \dfrac{1}{2}$, we get
${\sin ^{ - 1}}\dfrac{1}{2} \approx \dfrac{1}{2} + \dfrac{1}{2}\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^3}}}{3} + \dfrac{{1\cdot3}}{{2\cdot4}}\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^5}}}{5} + \dfrac{{1\cdot3\cdot5}}{{2\cdot4\cdot6}}\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^7}}}{7} + \dfrac{{1\cdot3\cdot5\cdot7}}{{2\cdot4\cdot6\cdot8}}\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^9}}}{9} \approx 0.523585$
Compare this result with the calculator value: ${\sin ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{6} \approx 0.523599$, the error is less than ${10^{ - 4}}$.
