Taylor's Theorem: 4th-deg. Maclaurin polynomial for cos(x)

[f1player]

Show, using Taylor’s Theorem, that the 4th degree Maclaurin polynomial for cos(x) can approximate cos(pi/6) correct to three decimal places.

Here's what i got:

The Maclaurin Polynomial is : 1 - 0.5x^2 + 1/24(x^4)
Subbing in pi/6 for x gives an approximation of 0.8660538834

From Taylor's Theorem: cos(pi/6) = Maclaurin Polynomial + error term

Error Term must equal < 0.0005 to be correct to 3 decimal places (is this correct?)

Now error Term = 1/(n+1)! *(pi/6)^(5) which equals 0.0003279 which is less than 0.0005

however the answer is given as 0.000163... < 0.0005

where did i go wrong?

 

[Opalg]

Re: Taylor's Theorem

Show, using Taylor’s Theorem, that the 4th degree Maclaurin polynomial for cos(x) can approximate cos(pi/6) correct to three decimal places.

Here's what i got:

The Maclaurin Polynomial is : 1 - 0.5x^2 + 1/24(x^4)
Subbing in pi/6 for x gives an approximation of 0.8660538834

From Taylor's Theorem: cos(pi/6) = Maclaurin Polynomial + error term

Error Term must equal < 0.0005 to be correct to 3 decimal places (is this correct?)

Now error Term = 1/(n+1)! *(pi/6)^(5) which equals 0.0003279 which is less than 0.0005

however the answer is given as 0.000163... < 0.0005

where did i go wrong?

If you are using the Lagrange form of the remainder then the error term is (1/5!)*(pi/6)^5*sin(y) for some y between 0 and pi/6. Then sin(y) =< sin(pi/6) = 1/2. That gives the answer 0.000163... .