Question

find the exact value.
cos\frac{11\pi}{6}
type + or -
\frac{\sqrt{}}{}

Explanation:

Step1: Use the cosine function property

The angle \(\frac{11\pi}{6}\) can be written as \(2\pi-\frac{\pi}{6}\).
We know that \(\cos(A - B)=\cos A\cos B+\sin A\sin B\), here \(A = 2\pi\), \(B=\frac{\pi}{6}\). Since \(\cos(2\pi)=1\) and \(\sin(2\pi)=0\), then \(\cos(2\pi-\frac{\pi}{6})=\cos(2\pi)\cos(\frac{\pi}{6})+\sin(2\pi)\sin(\frac{\pi}{6})=\cos(\frac{\pi}{6})\). Also, we know the unit - circle property \(\cos(\theta)=\cos(2\pi-\theta)\).

Step2: Find the value of \(\cos(\frac{\pi}{6})\)

From the unit - circle, for an angle \(\theta=\frac{\pi}{6}\) (or \(30^{\circ}\)), \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\). And since \(\frac{11\pi}{6}\) is in the fourth quadrant where the cosine function is positive.

Answer:

\(+\frac{\sqrt{3}}{2}\)