Question

The Maclaurin expansion of $\sin x$ is given by $\operatorname{Sin}x=\dfrac{x}{1!}-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-\dfrac{{{x}^{7}}}{7!}+.......$, where x is in radians. Use the series to compute the value of $\sin {{25}^{\circ }}$ with an accuracy of 0.001.

Answer 1

Hint: We start solving the problem by recalling the conversion of degrees to the radians. We then convert the given ${{25}^{\circ }}$ to the radians using this conversion. We then substitute the obtained value of radians in the place of x in the Maclaurin expansion. We then make the necessary calculations and neglect the values that were less than 0.001 to get the required result.

Complete step by step answer:
According to the problem, we have given Maclaurin expansion of $\sin x$ as $\operatorname{Sin}x=\dfrac{x}{1!}-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-\dfrac{{{x}^{7}}}{7!}+.......$, where x is in radians. We need to compute the value of $\sin {{25}^{\circ }}$ with an accuracy of 0.001 using this series.
Let us first convert ${{25}^{\circ }}$ into the radians. We know that ${{1}^{\circ }}$ is equal to $\dfrac{\pi }{180}$ radians or $0.0174$ radians. Using this we get ${{25}^{\circ }}=25\times 0.0174$.
$\Rightarrow {{25}^{\circ }}=0.435$ radians. Now, let us use this in the expansion of $\sin x$ to get the value of $\sin {{25}^{\circ }}$.
We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times .......$.
So, we have $\operatorname{Sin}{{25}^{\circ }}=\dfrac{0.435}{1}-\dfrac{{{\left( 0.435 \right)}^{3}}}{3\times 2\times 1}+\dfrac{{{\left( 0.435 \right)}^{5}}}{5\times 4\times 3\times 2\times 1}-\dfrac{{{\left( 0.435 \right)}^{7}}}{7\times 6\times 5\times 4\times 3\times 2\times 1}+.......$.
$\Rightarrow \operatorname{Sin}{{25}^{\circ }}=0.435-\dfrac{0.0823}{6}+\dfrac{0.0156}{120}-\dfrac{0.0029}{5040}+.......$.
$\Rightarrow \operatorname{Sin}{{25}^{\circ }}=0.435-0.014+0.00013$, we neglected next terms as we can see that the values are less than $0.001$.
$\Rightarrow \operatorname{Sin}{{25}^{\circ }}=0.421$.
So, we have found the approximate value of $\sin {{25}^{\circ }}$ as 0.421.

∴ The value of $\sin {{25}^{\circ }}$ computed by using Maclaurin expansion is 0.421.

Note: The value we got by solving using the Maclaurin method is an approximate value not the absolute value. To get the absolute value, we need to check the tables of the sine or we can check from the graph. The value we obtained will be as near as possible to the absolute value. We can also find the value of $\sin {{25}^{\circ }}$ from the \[\sin {{300}^{\circ }}\] by using $\sin 6\theta $ to get the absolute value.