2013th derivative of a trigonometric function Take the function
$f(x) = \sin(2x) \cdot \cos(x)$
Find the 2013 th  derivative
What I have found so far:
$f''(x)= -4\sin(x) \cdot \cos(2x) - 5f(x)$
I am assuming I need to find a relationship such as the above in order to just apply it again and again, however I can't seem to do so. 
Any help is appreciated. Thank You. 
=============================
EDIT: @Ihsan and Babak
Utilising this result, results in:
$f''(x) = -8\sin(3x) - f(x) $
And I'm not sure how to make recur this 1006 times!
 A: $$2\sin(2x)\cos(x)=\sin(3x)+\sin(x)$$ Use this.
A: Hint: Use $$\sin\alpha\cos β= \frac{1}2[\sin (\alpha + β) + \sin (\alpha − β)]$$ and $$\sin^{(n)}(x)=\frac{d^n}{dx^n}\sin(x)=\sin(x+\frac  {\pi}{2}n)$$ and use the Chain rule also.
A: Hint: Try to express the function using $e^x$ using: $\sin(\alpha) = \frac{e^{i \alpha} + e^{i\alpha}}{2i}$ and $\cos (\alpha) = \frac{e^{i\alpha}+e^{-i\alpha}}{2}$.
Full solution: (up to minor errors, if any...)
$$
\sin(2x) \cos(x) =\left(\frac{e^{i2x}-e^{-i2x}}{2i} \right)\left(\frac{e^{ix}+e^{-ix}}{2} \right)=\frac{1}{4i}(e^{i3x}-e^{-i3x}+e^{ix}-e^{-ix}).
$$ 
Since $(e^{ax})'=ae^{ax}$, the $2013$-th derivative is 
$$
(e^{ax})^{[2013]}=a^{2013}e^{ax}.
$$
Plugging this to the former equation and recalling that $i^{2013}=i^{2012}i=1^{503}i=i$, we get
$$
\begin{array}{lll}
(\sin(2x) \cos(x))^{[2013]}&=&\left(\frac{1}{4i}(e^{i3x}-e^{-i3x}+e^{ix}-e^{-ix})\right)^{[2013]}\\
&=&\frac{1}{4i} \big((i3)^{2013} e^{i3x}- (-i3)^{2013}e^{-i3x}+(i)^{2013} e^{ix}- (-i)^{2013}e^{-ix}\big)\\
&=&\frac{i}{4i}\big( 3^{2013}(e^{i3x}+e^{-i3x}) + e^{ix}+e^{-ix}\big)\\
&=&\frac{3^{2013}}{2}\frac{e^{i3x}+e^{-i3x}}{2} +\frac{1}{2}\frac{e^{ix}+e^{-ix}}{2}\\\
&=&\frac{3^{2013}}{2}\cos 3x+\frac{\cos x}{2}.
\end{array}
$$
A: Note the 2nd derivative of $\sin(a x)$ is $-a^2 \sin (a x)$ (and similar for $\cos a x)$ that is $\frac{d^2}{dx^2} f(a x) = -a^2 f(a x)$ when $f(x)$ is sine or cosine.  Doing two more derivatives you find the 4th derivative is $a^4 \sin( a x)$.  Repeating this four more times, you get the 8th derivative is $a^8 \sin (a x)$ as eight is a multiple of 4.  (Similarly 2012 is a multiple of 4; and then you have to do one more derivative to get to the 2013th derivative).
It helps to use trigonometric formulas to break the product of two trig functions into the sum of trig functions.
