Trigonometry (11th Edition) Clone

$$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=-1$$
$$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}$$ 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: $$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=\cos(\frac{7\pi}{9}+\frac{2\pi}{9})$$ $$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=\cos\frac{9\pi}{9}$$ $$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=\cos\pi$$ $$\cos\frac{7\pi}{9}\cos\frac{2\pi}{9}-\sin\frac{7\pi}{9}\sin\frac{2\pi}{9}=-1$$