Questions: -(pi/6)

Solution Steps

We start with the trigonometric identity for cosine of a difference: \[ \cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b) \]

We apply this identity to evaluate \(\cos\left(-\frac{\pi}{6}\right)\) by expressing it as: \[ \cos\left(-\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6} - \frac{\pi}{3}\right) \]

Using the identity, we can expand this as: \[ \cos\left(-\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{3}\right) \]

We substitute the known values of the trigonometric functions: \[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

Substituting these values into the equation gives: \[ \cos\left(-\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \] This simplifies to: \[ \cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \]

Final Answer