Question 9
Find the 3rd order Taylor Polynomial that approximates the below function near $x=0$.
\[
y=\sin 2 x
\]
a. $x-\frac{1}{3} x^{3}$
b. $x+\frac{1}{3} x^{3}$
c. $2 x-\frac{4}{3} x^{3}$
d. $2 x+\frac{4}{3} x^{3}$

Step 1 ：First, we need to recall the Taylor series expansion for the sine function, which is given by \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\).

Step 2 ：Since we are asked to find the 3rd order Taylor Polynomial, we only need to consider the terms up to \(x^3\). So, \(\sin x \approx x - \frac{x^3}{3!}\).

Step 3 ：However, we are asked to approximate \(\sin 2x\), not \(\sin x\). So, we substitute \(2x\) into the Taylor series expansion for \(\sin x\).

Step 4 ：This gives us \(\sin 2x \approx 2x - \frac{(2x)^3}{3!} = 2x - \frac{8x^3}{6} = 2x - \frac{4}{3}x^3\).

Step 5 ：So, the 3rd order Taylor Polynomial that approximates \(\sin 2x\) near \(x=0\) is \(\boxed{2x - \frac{4}{3}x^3}\).