How can you find the taylor expansion of #f(x) =sinx# about x=pi/6?

Answer 1

#sinx = sum_(n=0)^oo sin(((3n+1)pi)/6)/(n!)(x-pi/6)^n#

The general expression of the Taylor series of #f(x)# around #x=a# is:
#f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n#
To find the Taylor series of #sinx# we must therefore evaluate the derivatives of the function for all orders. The function #sinx# is indeed differentiable indefinitely in #RR#, and to obtain a synthetic notation for its derivatives we can use the following trigonometric equality:
#sin(x+pi/2) = cosx#

So we have:

#d/dx sinx = cosx = sin(x+pi/2)#
#d^2/dx^2 sinx = d/dx sin(x+pi/2) = cos(x+pi/2) = sin(x+pi)#

and clearly in general:

#d^n/dx^n sinx = sin(x+(npi)/2)#
For #x=pi/6# we have:
#[d^n/dx^n sinx]_(x=pi/6) = sin(pi/6+(npi)/2) = sin(((3n+1)pi)/6) #

The Taylor series is then:

#sinx = sum_(n=0)^oo sin(((3n+1)pi)/6)/(n!)(x-pi/6)^n#
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Answer 2

To find the Taylor expansion of ( f(x) = \sin(x) ) about ( x = \frac{\pi}{6} ), follow these steps:

  1. Determine the derivatives of ( f(x) = \sin(x) ) up to the desired order.
  2. Evaluate these derivatives at ( x = \frac{\pi}{6} ).
  3. Write the Taylor expansion using the derivatives and their respective coefficients.

Here are the detailed steps:

  1. Derivatives of ( f(x) = \sin(x) ):

    • First derivative: ( f'(x) = \cos(x) )
    • Second derivative: ( f''(x) = -\sin(x) )
    • Third derivative: ( f'''(x) = -\cos(x) )
    • Fourth derivative: ( f^{(4)}(x) = \sin(x) )
    • Fifth derivative: ( f^{(5)}(x) = \cos(x) )
    • Sixth derivative: ( f^{(6)}(x) = -\sin(x) )
    • Seventh derivative: ( f^{(7)}(x) = -\cos(x) )
    • Eighth derivative: ( f^{(8)}(x) = \sin(x) )
    • Continuing in this pattern, the derivatives alternate between sine and cosine functions.
  2. Evaluate the derivatives at ( x = \frac{\pi}{6} ):

    • ( 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} )
  3. Write the Taylor expansion using the derivatives and their respective coefficients:

    • ( \sin(x) = f\left(\frac{\pi}{6}\right) + f'\left(\frac{\pi}{6}\right)(x - \frac{\pi}{6}) + \frac{f''\left(\frac{\pi}{6}\right)}{2!}(x - \frac{\pi}{6})^2 + \frac{f'''\left(\frac{\pi}{6}\right)}{3!}(x - \frac{\pi}{6})^3 + \cdots )

Substitute the evaluated derivatives into the expansion to get the Taylor series expansion of ( \sin(x) ) about ( x = \frac{\pi}{6} ).

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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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