Q:

How would I calculate the coefficient for this taylor expansion? 

How would I find the coefficient of $x^{22}$ in the Taylor Series $\cos\bigg(\displaystyle {2x+\frac{\pi}{6}}\bigg)$ at $x=0$. I am aware on how to do Taylor expansion using the Maclaurin formula as well as the Taylor equation. 

A:

Note the following:
1) The derivatives of $\cos$ are periodic in the sense that $\cos(x)' = -\sin(x)$, $\cos(x)'' = -\cos(x)$, $\cos(x)''' = \sin(x)$ and $\cos(x)^{(4)} = \cos(x)$.
2) Every time you differentiate, you'll get an extra factor $2$ from the $2x+\frac{\pi}{6}$ inside the $\cos$ - this comes from the chain rule. 
Since $22 = 5\cdot 4 + 2$, differentiating $22$ times will give you a $-\cos$; and you will get $22$ factors $2$ outside the $\cos$. Hence:
$$\frac{d^{22}}{dx^{22}} \cos\left(2x+\frac{\pi}{6}\right) = -2^{22}\cos\left(2x+\frac{\pi}{6}\right)$$
To find the coefficient in the Taylor expansion at $x=0$, you then have to evaluate at $x=0$ and divide by $22!$. Since $\cos(\pi/6) = \sqrt{3}/2$, you get
$$\frac{-2^{21}\sqrt{3}}{22!}$$

