Answer
$$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=-1$$
Work Step by Step
$$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}$$
Remember the identity cosine of a sum:
$$\cos(A+B)=\cos A\cos B-\sin A\sin B$$
So we see here that $A$ resembles $\frac{7\pi}{9}$ and $B$ resembles $\frac{2\pi}{9}$.
That means we can shorten the exercise as follows:
$$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=\cos(\frac{7\pi}{9}+\frac{2\pi}{9})$$
$$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=\cos\frac{9\pi}{9}$$
$$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=\cos\pi$$
$$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=-1$$
