Question

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

Answer 1

`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(x)=.523599-.02392+.000328-2.1xx10^-6+8.15xx10-9~~.5`
Using Calculator: `=>sin(pi/6)=.5`

Explanation:

Solution Strategy:
Use the definition of Taylor series for a function, `f(x)` 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(x)` 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
`(df(x))/(dx)|_(x=0) = cos(0) =1`

`(d^(2)f(x))/(dx)^2|_(x=a) = -sin(0) =0`

`(d^(3)f(x))/(dx)^3|_(x=a) = -cos(0) = -1`

`(d^(4)f(x))/(dx)^3|_(x=a) = -sin(0) = 0`

`(d^(5)f(x))/(dx)^5|_(x=a) = cos(0) = 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(x)` at `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(x)=.523599-.02392+.000328-2.1xx10^-6+8.15xx10-9~~.5`
Using Calculator: `=>sin(pi/6)=.5`

It is likely that your calculator is using Taylor or likely MAcLaurin series to compute, `sin(x)`

Answer 2

`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 = pi/6`, rather than evaluating the Maclaurin series for `sin(x)`.

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

First, calculate the first few derivatives of `f(x) = sin(x)`.

`f(x) = sin(x)`
`f'(x) = cos(x)`
`f''(x) = -sin(x)`
`f'''(x) = -cos(x)`
`f^(4)(x) = sin(x)`

Since `f^((4))(x)` is the same as `f(x)`, the derivatives of `sin(x)` will continue to cycle like this forever.

Now, we plug in `pi/6` to these derivatives.

`f(pi/6) = sin(pi/6) = 1/2`
`f'(pi/6) = cos(pi/6) = sqrt3/2`
`f''(pi/6) = -sin(pi/6)= -1/2`
`f'''(pi/6) = -cos(pi/6)= -sqrt3/2`
`f^((4))(pi/6) = sin(pi/6) = 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(x)` (remember to start with term 0, using `f^((0))(a) = f(a)`):

`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