attempt 2: the answer is incorrect. 8 attempts remaining. use the definition of a taylor series to find the…

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ centered at $a$ is given by $\sum_{k = 0}^{\infty}\frac{f^{(k)}(a)}{k!}(x - a)^{k}$, where $f^{(k)}(a)$ is the $k$-th derivative of $f(x)$ evaluated at $x=a$. First, find the derivatives of $y = f(x)=\cos(6x)$. The $n$-th derivative of $\cos(6x)$ is $f^{(n)}(x)=6^{n}\cos(6x+\frac{n\pi}{2})$.

Step2: Evaluate derivatives at $a=\frac{\pi}{12}$

Let $a = \frac{\pi}{12}$. Then $f^{(n)}(\frac{\pi}{12})=6^{n}\cos(6\times\frac{\pi}{12}+\frac{n\pi}{2})=6^{n}\cos(\frac{\pi}{2}+\frac{n\pi}{2})$. When $n = 2k$ (even), $f^{(2k)}(\frac{\pi}{12})=6^{2k}\cos(\frac{\pi}{2}+k\pi)=0$. When $n = 2k + 1$ (odd), $f^{(2k + 1)}(\frac{\pi}{12})=6^{2k+1}\cos(\frac{\pi}{2}+(2k + 1)\frac{\pi}{2})=(- 1)^{k}6^{2k+1}$.

Step3: Write out the Taylor - series

Substitute into the Taylor - series formula. Since $f^{(2k)}(\frac{\pi}{12}) = 0$, the non - zero terms come from odd - numbered derivatives. The Taylor series is $\sum_{k = 0}^{\infty}\frac{f^{(2k+1)}(\frac{\pi}{12})}{(2k + 1)!}(x-\frac{\pi}{12})^{2k+1}=\sum_{k = 0}^{\infty}\frac{(-1)^{k}6^{2k+1}}{(2k + 1)!}(x-\frac{\pi}{12})^{2k+1}$.

Answer:

$\sum_{k = 0}^{\infty}\frac{(-1)^{k}6^{2k+1}}{(2k + 1)!}(x-\frac{\pi}{12})^{2k+1}$