a.) Find the Taylor polynomial of degree 3, expanded about a=0, for f(x)= sin(2x).

I got P3(x)= 2x-8/3!(x)^3, which is correct.

What I don't understand is this part:
b.) Use your answer from part a.) to approximate sin(2/3)

So to do this I just plugged in 2/3 into x, like this:

sin(2/3)= (2/3)-8/3!(2/3)^3

BUT my instructor crossed out the 2's on top and replaced it with 1's. Why is that?

Question

a.) Find the Taylor polynomial of degree 3, expanded about a=0, for f(x)= sin(2x).

I got P3(x)= 2x-8/3!(x)^3, which is correct.

What I don't understand is this part:
b.) Use your answer from part a.) to approximate sin(2/3)

So to do this I just plugged in 2/3 into x, like this:

sin(2/3)= (2/3)-8/3!(2/3)^3

BUT my instructor crossed out the 2's on top and replaced it with 1's. Why is that?

anonymous · Answer

$$P _{3}^{}(x)= 2x-8/3!(x)^3$$

anonymous · Answer

$$\sin(2/3)= 2(2/3)-8/3!(2/3)^3$$

anonymous · Answer

The Taylor polynomial you found is the expansion of sin(2x).  If you want sin(2/3), you need to replace substitute x= 1/3.

anonymous · Answer

you approximated sin(2x) so x=1/3 to approximate sin(2/3)