Question

How do you find the Taylor series of $f(x) = \sin x$ at $a = \dfrac{\pi }{6}$?

Answer 1

Hint: Here we have to find the given function at the point by using the Taylor series expansion. Then we differentiate the given term and put the taylor formula. After doing some simplification we get the final answer.

Complete step-by-step solution:
For the given question, we will use the definition of taylor series.
We know taylor series for a function $f(x)$ is given by:
$f(x) = f(a) + f'(a)\dfrac{{x - a}}{1}! + f''(a)\dfrac{{{{(x - a)}^2}}}{{2!}} + f'''(a)\dfrac{{{{(x - a)}^3}}}{3}! + .... + {f^n}(a)\dfrac{{{{(x - a)}^n}}}{{n!}}$
Therefore, the taylor series can also be written as:
\[f(x) = {f^n}(a)\dfrac{{{{(x - a)}^n}}}{{n!}}\]
Now we have $f(x) = \sin x$, we will find the derivatives of the function at $x = 0$ to substitute it in the taylor series.
Now we will find the derivatives of $f(x) = \sin (x)$.
Now the first derivative of $f(x) = \sin (x)$ is:
$f'(x) = \dfrac{{dy(f(x))}}{{dx}}{|_{x = 0}} = \cos (0) = 1$
The second derivative of $f(x) = \sin (x)$ is:
$f''(x) = \dfrac{{{d^2}y(f(x))}}{{d{x^2}}}{|_{x = 0}} = - \sin (0) = 0$
The third derivative of $f(x) = \sin (x)$ is:
${f^3}(x) = \dfrac{{{d^3}y(f(x))}}{{d{x^3}}}{|_{x = 0}} = - \cos (0) = - 1$
The fourth derivative of $f(x) = \sin (x)$ is:
${f^4}(x) = \dfrac{{{d^4}y(f(x))}}{{d{x^4}}}{|_{x = 0}} = - \sin (0) = 0$
The fifth derivative of $f(x) = \sin (x)$ is:
${f^5}(x) = \dfrac{{{d^5}y(f(x))}}{{d{x^5}}}{|_{x = 0}} = \cos (0) = 1$
Now since this will keep on recurring forever and we can see that all the derivatives which even have the value $0$ and all the odd values will toggle between $1$ and $ - 1$ which will keep on recurring.
Thus, we can write the equation as:
$\sin (x) = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} + \dfrac{{{x^7}}}{{7!}} + ....$ for all $x$
Now we will evaluate $f(x) = \sin (x)$ at $\dfrac{\pi }{6}$ in the first $5$terms:
$\Rightarrow$$\sin {(x)_{x = \dfrac{\pi }{6}}} = \dfrac{\pi }{6} - \dfrac{{{{\left( {\dfrac{\pi }{6}} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {\dfrac{\pi }{6}} \right)}^5}}}{{5!}} - \dfrac{{{{\left( {\dfrac{\pi }{6}} \right)}^7}}}{{7!}} + \dfrac{{{{\left( {\dfrac{\pi }{6}} \right)}^9}}}{{9!}}$
Now using the calculator and simplifying the values we can simply the above equation as:
$\Rightarrow$$\sin (x) = 0.5235 - 0.2392 + 0.0003 - 2.1 \times {10^{ - 6}} + 8.15 \times 10 - 9$
On simplifying we get:
$\Rightarrow$$\sin (x) \approx 0.5$

Therefore, the taylor expansion of $f(x) = \sin (x)$ at $\dfrac{\pi }{6}$ is approximately equal to $0.5$.

Note: By using the taylor series the answer we get is an approximate answer and not the absolute value, the absolute answer when calculated from a calculator might vary.
To cross check whether the answer is correct, the original function should be calculated by using a calculator and if the answer is close to the answer which we get from the taylor series, then the answer is correct.
The value of $\sin \left( {\dfrac{\pi }{6}} \right) = 0.5$ when calculated from a calculator.