what is the taylor series of f(x) = sin(x) at a=pi/3

Answer 1

In google paste:

what is the Taylor series of f(x) = sin(x) at a=pi/3

When you see list of results go on:

How do you use Taylor series for sin(x) at a = pi/3? | Socratic

You will see solution with explanation.

Answer 2

sinx = x - x^3/3! + x^5/5! - x^7/7! + ..... for x = π/3
= π/3 - (π/3)^3/6 + (π/3)^5/120 - (π/3)^7/5040 + ...
= 1.047197... - .191396... + .010494... - .000274012 + .... (the next < 3 decimals)
= appr .86602127

real answer : sin π/3 = .866025403 correct to 9 decimals
error : .000004132 , not bad

Answer 3

Reiny

Your answer is Maclaurin series, but here it is need Taylor series.

Of course Maclaurin series is a Taylor series expansion of a function about 0.

Here it is need Taylor series of a function about π / 3 ≈ 1

Your answer is also correct, but I think something else is answer for this question.

Answer 4

You are right, I skimmed over the "Taylor" part.
I should read more carefully.