Answer
$$\frac{d^{999}}{dx^{999}}(\cos x)=\sin x$$
Work Step by Step
We have $$\frac{d}{dx}(\cos x)=(\cos x)'=-\sin x$$
$$\frac{d^2}{dx^2}(\cos x)=(-\sin x)'=-\cos x$$
$$\frac{d^3}{dx^3}(\cos x)=(-\cos x)'=-(-\sin x)=\sin x$$
$$\frac{d^4}{dx^4}(\cos x)=(\sin x)'=\cos x$$
This ends the first cycle.
$$\frac{d^5}{dx^5}(\cos x)=(\cos x)'=-\sin x$$
$$\frac{d^6}{dx^6}(\cos x)=(-\sin x)'=-\cos x$$
$$\frac{d^7}{dx^7}(\cos x)=(-\cos x)'=-(-\sin x)=\sin x$$
$$\frac{d^8}{dx^8}(\cos x)=(\sin x)'=\cos x$$
This ends the second cycle.
We can generalize the pattern as follows:
$\frac{d^{4n+1}}{dx^{4n+1}}(\cos x)=-\sin x$, $\frac{d^{4n+2}}{dx^{4n+2}}(\cos x)=-\cos x$, $\frac{d^{4n+3}}{dx^{4n+3}}(\cos x)=\sin x$ and $\frac{d^{4n}}{dx^{4n}}(\cos x)=\cos x$ (for $n\in N$)
Here $999=4n+3$, therefore, $$\frac{d^{999}}{dx^{999}}(\cos x)=\sin x$$
