use a half - angle formula to find the exact value of cos 5π/12. cos 5π/12 =

Explanation:

Step1: Recall the half - angle formula

$\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1 + \cos\alpha}{2}}$. We know that $\frac{5\pi}{12}=\frac{\frac{5\pi}{6}}{2}$, so $\alpha=\frac{5\pi}{6}$.

Step2: Determine the sign

Since $\frac{\pi}{2}<\frac{5\pi}{12}<\pi$, $\cos\frac{5\pi}{12}<0$.

Step3: Find $\cos\alpha$

$\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$.

Step4: Substitute into the formula

$\cos\frac{5\pi}{12}=-\sqrt{\frac{1+\cos\frac{5\pi}{6}}{2}}=-\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=-\sqrt{\frac{2 - \sqrt{3}}{4}}=-\frac{\sqrt{2-\sqrt{3}}}{2}$. But we can also rewrite it in a more standard form. We know that $\cos\frac{5\pi}{12}=\frac{\sqrt{2-\sqrt{3}}}{2}$ (taking the positive value as we usually consider the principal - value form for exact trigonometric values in this context). Another way: We know that $\cos(A - B)=\cos A\cos B+\sin A\sin B$. $\frac{5\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6}$. $\cos\frac{5\pi}{12}=\cos(\frac{\pi}{4}+\frac{\pi}{6})=\cos\frac{\pi}{4}\cos\frac{\pi}{6}-\sin\frac{\pi}{4}\sin\frac{\pi}{6}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

Answer:

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