you can retry this question below\nfind the exact value of 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 $\cos(\frac{19\pi}{12})=\cos(\pi+\frac{7\pi}{12})$. According to the cosine - addition formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$, when $A = \pi$ and $B=\frac{7\pi}{12}$, we know that $\cos(\pi+\theta)=-\cos\theta$, so $\cos(\pi+\frac{7\pi}{12})=-\cos(\frac{7\pi}{12})$. And $\frac{7\pi}{12}=\frac{3\pi}{12}+\frac{4\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3}$.

Step2: Apply the cosine - addition formula

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

Step3: Get the final result

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:

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