Exercise 9

Recall that we found in Preview and subsequent work that $\sin(x) \approx x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7$, which is the degree $7$ Taylor approximation centered at $0$. And in Activity, we found that the degree $6$ Taylor approximation centered at $a = 0$ for $\cos(x)$ is $\cos(x) \approx 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6$.

In this exercise, we investigate Taylor polynomial approximations of $f(x) = \sin(x)$ centered at $a = \frac{\pi}{2}$.

1. By finding the appropriate derivatives of $f(x) = \sin(x)$ and evaluating them at $a = \frac{\pi}{2}$, determine the degree $6$ Taylor polynomial approximation of $\sin(x)$ centered at $a = \frac{\pi}{2}$.

2. How is your result in (a) similar to the degree $6$ Taylor polynomial of $\cos(x)$ that is centered at $a = 0$?

3. Recall the trigonometric identity that states $\sin(x) = \cos(x - \frac{\pi}{2})$. How does this identity help explain what you found in (a) and (b)?