Problem 34

To solve the problem step-by-step, we start with the given equation and apply trigonometric identities and properties to find the value of $\cos(2\alpha + \frac{2\pi}{3})$.

1. Given: $\sin(\alpha - \frac{\pi}{6}) = \frac{2}{3}$.

2. Use the identity $\sin(x) = -\cos(\frac{\pi}{2} + x)$ to rewrite the given equation:
$$\sin(\alpha - \frac{\pi}{6}) = -\cos\left(\frac{\pi}{2} + (\alpha - \frac{\pi}{6})\right) = -\cos\left(\alpha + \frac{\pi}{3}\right) = \frac{2}{3}.$$
This implies that $\cos\left(\alpha + \frac{\pi}{3}\right) = -\frac{2}{3}$.

3. To find $\cos(2\alpha + \frac{2\pi}{3})$, we use the double angle formula for cosine, $\cos(2x) = 2\cos^2(x) - 1$:
$$\cos(2\alpha + \frac{2\pi}{3}) = 2\cos^2\left(\alpha + \frac{\pi}{3}\right) - 1.$$
Substituting $\cos\left(\alpha + \frac{\pi}{3}\right) = -\frac{2}{3}$ into the equation:
$$\cos(2\alpha + \frac{2\pi}{3}) = 2\left(-\frac{2}{3}\right)^2 - 1 = 2\left(\frac{4}{9}\right) - 1 = \frac{8}{9} - 1 = -\frac{1}{9}.$$

Therefore, the final answer is $\boxed{-\frac{1}{9}}$.