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How do you write 5th degree taylor polynomial for sin(x)?

Answer 1

With both x and a in radian measure,
#sin x =sin a+(x-a)cos a-(x-a)^2/(2!)sin a -(x-a)^3/(3!)cos a+(x-a)^4/(4!)sina+(x-a)^5/(5!)cosa +O(x-a)^6#

Taylor expansion is about a neighboring point x = a, in contrast to

Maclaurin's that is about x = 0.

The expansion is

f(x) f(a) +(x-a)f'(a)+(x-a)^2/(2!)f''(a)+(x-a)^3/(3!)f'''(a)+.... Accordingly, here

#sin x =sin a+(x-a)cos a-(x-a)^2/(2!)sin a -(x-a)^3/(3!)cos a#

#+(x-a)^4/(4!)sina+(x-a)^5/(5!)cosa +O(x-a)^6#

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Answer 2

Axel Alvarez

The fifth-degree Taylor polynomial for sin(x) is given by:

[ P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} ]

This polynomial approximates the function sin(x) near the point x = 0 up to the fifth-degree term.

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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.