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

1 Answer
Sep 15, 2015

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 )