Find the 5th Taylor polynomial for f (x) = sin x about x0 = 0

To find the 5th Taylor polynomial for f(x) = sin(x) about x0 = 0, we can use the formula for the nth term of the Taylor series expansion:

Tn(x) = f(x0) + f'(x0)(x - x0)/1! + f''(x0)(x - x0)^2/2! + ... + f^n(x0)(x - x0)^n/n!

First, let's find the values of f(x0), f'(x0), f''(x0), and so on, up to the 5th derivative.

f(x) = sin(x) f(x0) = sin(0) = 0

f'(x) = cos(x) f'(x0) = cos(0) = 1

f''(x) = -sin(x) f''(x0) = -sin(0) = 0

f'''(x) = -cos(x) f'''(x0) = -cos(0) = -1

f''''(x) = sin(x) f''''(x0) = sin(0) = 0

f'''''(x) = cos(x) f'''''(x0) = cos(0) = 1

Now, let's substitute these values into the formula for the 5th Taylor polynomial:

T5(x) = f(0) + f'(0)(x - 0)/1! + f''(0)(x - 0)^2/2! + f'''(0)(x - 0)^3/3! + f''''(0)(x - 0)^4/4! + f'''''(0)(x - 0)^5/5!

T5(x) = 0 + 1(x)/1! + 0(x^2)/2! + (-1)(x^3)/3! + 0(x^4)/4! + 1(x^5)/5!

Simplifying the terms:

T5(x) = x - (x^3)/6 + (x^5)/120

Therefore, the 5th Taylor polynomial for f(x) = sin(x) about x0 = 0 is T5(x) = x - (x^3)/6 + (x^5)/120.