find the exact value. cos(19π/12)

Explanation:

Step1: Rewrite the angle

We can write $\frac{19\pi}{12}=\frac{12\pi + 7\pi}{12}=\pi+\frac{7\pi}{12}$. Then $\cos(\frac{19\pi}{12})=\cos(\pi+\frac{7\pi}{12})$. Using the cosine - addition formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$ with $A=\pi$ and $B = \frac{7\pi}{12}$, and knowing that $\cos(\pi)=- 1$ and $\sin(\pi)=0$, we get $\cos(\pi+\frac{7\pi}{12})=-\cos(\frac{7\pi}{12})$. Also, $\frac{7\pi}{12}=\frac{3\pi}{12}+\frac{4\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3}$.

Step2: Apply the cosine - addition formula

The cosine - addition formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Here $A=\frac{\pi}{4}$ and $B=\frac{\pi}{3}$. We know that $\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, $\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, $\cos(\frac{\pi}{3})=\frac{1}{2}$, and $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$. So $\cos(\frac{\pi}{4}+\frac{\pi}{3})=\cos(\frac{\pi}{4})\cos(\frac{\pi}{3})-\sin(\frac{\pi}{4})\sin(\frac{\pi}{3})=\frac{\sqrt{2}}{2}\times\frac{1}{2}-\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}=\frac{\sqrt{2}-\sqrt{6}}{4}$.

Step3: Find the final value

Since $\cos(\frac{19\pi}{12})=-\cos(\frac{7\pi}{12})$ and $\cos(\frac{7\pi}{12})=\frac{\sqrt{2}-\sqrt{6}}{4}$, then $\cos(\frac{19\pi}{12})=-\frac{\sqrt{2}-\sqrt{6}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

Answer:

$\frac{\sqrt{6}-\sqrt{2}}{4}$