Answer
$$\cos(-24^\circ)=\cos16^\circ-\cos40^\circ$$
The statement is false.
Work Step by Step
$$\cos(-24^\circ)=\cos16^\circ-\cos40^\circ$$
We can rewrite $-24^\circ$ as follows, $$-24^\circ=16^\circ-40^\circ$$
That means $$\cos(-24^\circ)=\cos(16^\circ-40^\circ)$$
Using the cosine difference identity, which states
$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$
we can expand $\cos(16^\circ-40^\circ)$
$$\cos(-24^\circ)=\cos16^\circ\cos40^\circ+\sin16^\circ\sin40^\circ$$
And since $\cos16^\circ\cos40^\circ+\sin16^\circ\sin40^\circ\ne\cos16^\circ-\cos40^\circ$, that means $$\cos(-24^\circ)\ne\cos16^\circ-\cos40^\circ$$
The statement is false.
Until now, it also should be clear to you that $\cos(16^\circ-40^\circ)\ne\cos16^\circ-\cos40^\circ$.
