## Trigonometry (11th Edition) Clone

The statement $$\cos\frac{2\pi}{3}=\cos\frac{11\pi}{12}\cos\frac{\pi}{4}+\sin\frac{11\pi}{12}\sin\frac{\pi}{4}$$ is true.
$$\cos\frac{2\pi}{3}=\cos\frac{11\pi}{12}\cos\frac{\pi}{4}+\sin\frac{11\pi}{12}\sin\frac{\pi}{4}$$ We now try subtracting $\frac{\pi}{4}$ from $\frac{11\pi}{12}$, $$\frac{11\pi}{12}-\frac{\pi}{4}=\frac{11\pi}{12}-\frac{3\pi}{12}=\frac{8\pi}{12}=\frac{2\pi}{3}$$ Hence, we can rewrite $\cos\frac{2\pi}{3}$: $$\cos\frac{2\pi}{3}=\cos\Big(\frac{11\pi}{12}-\frac{\pi}{4}\Big)$$ Now we use the cosine difference identity $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ (be extremely careful about the sign in the middle) to expand $\cos\Big(\frac{11\pi}{12}-\frac{\pi}{4}\Big)$: $$\cos\frac{2\pi}{3}=\cos\frac{11\pi}{12}\cos\frac{\pi}{4}+\sin\frac{11\pi}{12}\sin\frac{\pi}{4}$$ This means that the statement $$\cos\frac{2\pi}{3}=\cos\frac{11\pi}{12}\cos\frac{\pi}{4}+\sin\frac{11\pi}{12}\sin\frac{\pi}{4}$$ is true.