## Trigonometry (11th Edition) Clone

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