How would I calculate the coefficient for this taylor expansion?

Question

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.

Answer 1

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 $$\frac{-2^{21}\sqrt{3}}{22!}$$

Answer 2

With some trig identities, you should notice that

$$\cos\left(2x+\frac\pi6\right)=\frac{\sqrt3}2\cos(2x)+\frac12\sin(2x)$$

Using known Taylor expansions,

$$\cos(2x)=\sum_{k=0}^\infty\frac{(-1)^k(2x)^{2k}}{(2k)!}$$

Since $\sin(2x)$ will only have odd exponents, we can ignore it, leaving

$$\frac{-\sqrt3(2^{21})}{22!}x^{22}$$

Answer 3

$cos (2x + \frac \pi6) = \cos 2x \cos\frac \pi6 - \sin 2x \sin \frac \pi6$

the coefficient of the $x^{22}$ term is $(\cos\frac \pi6) \frac {(-1)^{11}2^{22}}{22!}$