Question

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`

Explanation:

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`