# How do you find the taylor series of f(x)=sinx at a=pi/6?

##### 2 Answers
Mar 30, 2016

sin(x) = x-x^3/(3!) +x^5/(5!)-x^7/(7!) + cdots  for all $x$
sin(x)=sum_(n=0)^oo (-1)^n/((2n+1)!)x^(2n+1)

sin(x)_(x=(pi/6)) = pi/6-(pi/6)^3/(3!)+(pi/6)^5/(5!)-(pi/6)^7/(7!)+(pi/6)^9/(9!)

$\sin \left(x\right) = .523599 - .02392 + .000328 - 2.1 \times {10}^{-} 6 + 8.15 \times 10 - 9 \approx .5$
Using Calculator: $\implies \sin \left(\frac{\pi}{6}\right) = .5$

#### Explanation:

Solution Strategy:
Use the definition of Taylor series for a function, $f \left(x\right)$ given by:
f(x)=f(a) + f^'(a)(x-a)/1! +f^('')(a)(x-a)^2/2! +f^(3)(a)(x-a)^(3)/(3! )+cdots+ f^(n)(a)(x-a)^n/(n!) + cdots

1) f(x) = sum_(n=0)^oo f^(n)(a)(x-a)^n/(n!)
Expansion $f \left(x\right)$ around zero will yield
f(x) = sum_(n=0)^oo f^(n)(0)(x)^n/(n!) this McLaurin Series
a special case of Taylor series
2) Find the
$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} {|}_{x = 0} = \cos \left(0\right) = 1$

$\frac{{d}^{2} f \left(x\right)}{\mathrm{dx}} ^ 2 {|}_{x = a} = - \sin \left(0\right) = 0$

$\frac{{d}^{3} f \left(x\right)}{\mathrm{dx}} ^ 3 {|}_{x = a} = - \cos \left(0\right) = - 1$

$\frac{{d}^{4} f \left(x\right)}{\mathrm{dx}} ^ 3 {|}_{x = a} = - \sin \left(0\right) = 0$

$\frac{{d}^{5} f \left(x\right)}{\mathrm{dx}} ^ 5 {|}_{x = a} = \cos \left(0\right) = 1$
$\vdots$

We see that all even derivatives are zero and all odd derivative toggle between -1 and 1. Thus we write

sin(x) = x-x^3/(3!) +x^5/(5!)-x^7/(7!) + cdots  for all $x$
sin(x)=sum_(n=0)^oo (-1)^n/((2n+1)!)x^(2n+1)

Now let's evaluate $\sin \left(x\right)$ at $\frac{\pi}{6}$ for the first 5 terms:

sin(x)_(x=(pi/6)) = pi/6-(pi/6)^3/(3!)+(pi/6)^5/(5!)-(pi/6)^7/(7!)+(pi/6)^9/(9!)

$\sin \left(x\right) = .523599 - .02392 + .000328 - 2.1 \times {10}^{-} 6 + 8.15 \times 10 - 9 \approx .5$
Using Calculator: $\implies \sin \left(\frac{\pi}{6}\right) = .5$

It is likely that your calculator is using Taylor or likely MAcLaurin series to compute, $\sin \left(x\right)$

Mar 26, 2017

sin(x) = sum_(n=0)^oo (-1)^n(((x-pi/6)^(2n))/(2*(2n)!) + (sqrt3(x-pi/6)^(2n+1))/(2*(2n+1)!))

#### Explanation:

I believe the question is asking for a Taylor series centered around $a = \frac{\pi}{6}$, rather than evaluating the Maclaurin series for $\sin \left(x\right)$.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

First, calculate the first few derivatives of $f \left(x\right) = \sin \left(x\right)$.

$f \left(x\right) = \sin \left(x\right)$
$f ' \left(x\right) = \cos \left(x\right)$
$f ' ' \left(x\right) = - \sin \left(x\right)$
$f ' ' ' \left(x\right) = - \cos \left(x\right)$
${f}^{4} \left(x\right) = \sin \left(x\right)$

Since ${f}^{\left(4\right)} \left(x\right)$ is the same as $f \left(x\right)$, the derivatives of $\sin \left(x\right)$ will continue to cycle like this forever.

Now, we plug in $\frac{\pi}{6}$ to these derivatives.

$f \left(\frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$
$f ' \left(\frac{\pi}{6}\right) = \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
$f ' ' \left(\frac{\pi}{6}\right) = - \sin \left(\frac{\pi}{6}\right) = - \frac{1}{2}$
$f ' ' ' \left(\frac{\pi}{6}\right) = - \cos \left(\frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$
${f}^{\left(4\right)} \left(\frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

The formula for the $n$th term of a Taylor polynomial around $x = a$ is:

f^((n))(a)/(n!)*(x-a)^n

So, we can write out the first few terms of our Taylor polynomial for $\sin \left(x\right)$ (remember to start with term 0, using ${f}^{\left(0\right)} \left(a\right) = f \left(a\right)$):

sin(x) = (1/2)/(0!) * (x-pi/6)^0+(sqrt3/2)/(1!) * (x-pi/6)^1 + (-1/2)/(2!) * (x-pi/6)^2 + (-sqrt3/2)/(3!) * (x-pi/6)^3 + (1/2)/(4!) * (x-pi/6)^4+...

Simplifying this a bit, we get:

sin(x) = 1/2 + (sqrt3(x-pi/6))/2 - (x-pi/6)^2/(2*2!) - (sqrt3(x-pi/6)^3)/(2*3!) + (x-pi/6)^4/(2*4!)+...

This is technically the final answer but the formal way to write it is with summation notation. If we group every two terms together, we can write this polynomial as:

sin(x) = sum_(n=0)^oo (-1)^n(((x-pi/6)^(2n))/(2*(2n)!) + (sqrt3(x-pi/6)^(2n+1))/(2*(2n+1)!))

And remember: don't be bogged down by how big or complex this summation is; it's just a set of instructions for writing out the polynomial we just made.

Final Answer