Question

How do you use Taylor series for `sin(x)` at `a = pi/3`?

Answer 1

Just plug values for `x` into it that are close to `pi/3` (like `x=1`) to get good estimates for `sin(x)` near such values of `x`. Truncating the series at higher powers gives better approximations for a given `x`.

Explanation:

Since `f(x)=sin(x)` implies that `f'(x)=cos(x)`, `f''(x)=-sin(x)`, `f'''(x)=-cos(x)`, `f''''(x)=sin(x)`, etc..., it follows that if `a=pi/3`, then `f(a)=sqrt(3)/2`, `f'(a)=1/2`, `f''(a)=-sqrt(3)/2`, `f'''(a)=-1/2`, `f''''(a)=sqrt(3)/2`, etc...

The Taylor series for `sin(x)` at `a=pi/3` is therefore:

`f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+(f''''(a))/(4!)(x-a)^4+cdots`

`=sqrt(3)/2+1/2(x-pi/3)-sqrt(3)/4(x-pi/3)^2-1/12(x-pi/3)^3+sqrt(3)/48(x-pi/3)^4+cdots`.

If you, for example, substitute `x=1` (radian) into this expansion and stop at the 4th powered term, you'll get:

`sin(1)approx sqrt(3)/2+1/2(1-pi/3)-sqrt(3)/4(1-pi/3)^2-1/12(1-pi/3)^3+sqrt(3)/48(1-pi/3)^4 approx 0.84147`

Using a calculator (in radian mode) gives the same approximation to 5 decimal places (actually, if you carry it out further, it's accurate to 10 decimal places).

To know how accurate you are guaranteed to be requires more theory (see especially Theorem 4 at the following link: http://www.stewartcalculus.com/data/CALCULUS%20Early%20Transcendentals/upfiles/Formulas4RemainderTaylorSeries5ET.pdf )