Finding the $2008$th derivative of $\sin^6(\frac{x}{4}) + \cos^6\left(\frac{x}{4}\right)$ at $x = 0$? 
Let $f(x) = \sin^6 \frac{x}{4} + \cos^6 \frac{x}{4}$ for all real numbers $x$. Determine $f^{(2008)}(0)$ (ie. $f$ differentiated $2008$ times and then evaluated at $x=0$)

Hi,
 I have tried to find the $2008$th derivative at $x = 0$ of this function and after finding the 4th derivative the equation is simply becoming too long and I feel like I am doing something wrong. Can someone help me please?
 A: Several among the list of trigonometric identities to consider are
$$\sin^2(x) = 1 - \cos^2(x) \tag{1}\label{eq1A}$$
$$\sin(2y) = 2\sin(y)\cos(y) \implies \sin(y)\cos(y) = \frac{\sin(2y)}{2} \tag{2}\label{eq2A}$$
$$\cos(2y) = 1 - 2\sin^2(y) \implies \sin^2(y) = \frac{1 - \cos(2y)}{2} \tag{3}\label{eq3A}$$
Using these identities, you have
$$\begin{equation}\begin{aligned}
f(x) & = \sin^6\left(\frac{x}{4}\right) + \cos^6\left(\frac{x}{4}\right) \\
& = \left(\sin^2\left(\frac{x}{4}\right)\right)^3 + \cos^6\left(\frac{x}{4}\right) \\
& = \left(1 - \cos^2\left(\frac{x}{4}\right)\right)^3 + \cos^6\left(\frac{x}{4}\right) \\
& = 1 - 3\cos^2\left(\frac{x}{4}\right) + 3\cos^4\left(\frac{x}{4}\right) - \cos^6\left(\frac{x}{4}\right) + \cos^6\left(\frac{x}{4}\right) \\
& = 1 - 3\cos^2\left(\frac{x}{4}\right)\left(1 - \cos^2\left(\frac{x}{4}\right)\right) \\
& = 1 - 3\cos^2\left(\frac{x}{4}\right)\sin^2\left(\frac{x}{4}\right) \\
& = 1 - 3\left(\frac{1}{2}\sin\left(\frac{x}{2}\right)\right)^2 \\
& = 1 - \frac{3}{4}\left(\sin^2\left(\frac{x}{2}\right)\right) \\
& = 1 - \frac{3}{4}\left(\frac{1 - \cos(x)}{2}\right) \\
& = 1 - \frac{3}{8} + \frac{3}{8}\cos(x) \\
& = \frac{5}{8} + \frac{3}{8}\cos(x)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This form is much easier to work with to determine your derivative and resulting value, which I'll leave you to do as it's almost trivial now.
