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#.

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: https://tutor.hix.ai )

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Answer 2

To use the Taylor series for ( \sin(x) ) centered at ( a = \frac{\pi}{3} ), you can expand ( \sin(x) ) around this point using its Taylor series formula:

[ \sin(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n ]

where ( f^{(n)}(a) ) represents the ( n )-th derivative of ( \sin(x) ) evaluated at ( x = a ).

First, find the derivatives of ( \sin(x) ):

[ f(x) = \sin(x) ] [ f'(x) = \cos(x) ] [ f''(x) = -\sin(x) ] [ f'''(x) = -\cos(x) ]

Now, evaluate these derivatives at ( x = \frac{\pi}{3} ):

[ f\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} ] [ f'\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} ] [ f''\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} ] [ f'''\left(\frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2} ]

Now, plug these values into the Taylor series formula:

[ \sin(x) \approx \frac{\sqrt{3}}{2} + \frac{1}{2}(x - \frac{\pi}{3}) - \frac{\sqrt{3}}{2}\frac{(x - \frac{\pi}{3})^2}{2!} - \frac{1}{2}\frac{(x - \frac{\pi}{3})^3}{3!} ]

This is the Taylor series for ( \sin(x) ) centered at ( a = \frac{\pi}{3} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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