We are given the function $y = a \sin \theta + b \cos \theta$ where $0 \le \theta < 2\pi$. The function has a maximum value of 2 at $\theta = \frac{2}{3}\pi$. We need to find the values of the constants $a$ and $b$, and then find the minimum value of $y$ and the corresponding value of $\theta$.

1. Problem Description

We are given the function $y = a \sin \theta + b \cos \theta$ where $0 \le \theta < 2\pi$. The function has a maximum value of 2 at $\theta = \frac{2}{3}\pi$.

We need to find the values of the constants $a$ and $b$, and then find the minimum value of $y$ and the corresponding value of $\theta$.

2. Solution Steps

(1) We are given that the maximum value of $y = a \sin \theta + b \cos \theta$ is 2 when $\theta = \frac{2\pi}{3}$. Thus,

$a \sin \left(\frac{2\pi}{3}\right) + b \cos \left(\frac{2\pi}{3}\right) = 2$.

Since $\sin \left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2}$, we have

$a \left(\frac{\sqrt{3}}{2}\right) + b \left(-\frac{1}{2}\right) = 2$, which simplifies to

$\sqrt{3}a - b = 4$.

We can rewrite the function as $y = R \sin(\theta + \alpha)$, where $R = \sqrt{a^2 + b^2}$ and $\alpha$ is an angle such that $\cos \alpha = \frac{a}{R}$ and $\sin \alpha = \frac{b}{R}$.

The maximum value of $y$ is $R = \sqrt{a^2 + b^2}$, which is given to be

2. Thus, $\sqrt{a^2 + b^2} = 2$, so $a^2 + b^2 = 4$.

Also, the maximum occurs when $\theta + \alpha = \frac{\pi}{2}$, so $\frac{2\pi}{3} + \alpha = \frac{\pi}{2}$, which implies $\alpha = \frac{\pi}{2} - \frac{2\pi}{3} = -\frac{\pi}{6}$.

Then, $\cos \alpha = \cos \left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} = \frac{a}{R} = \frac{a}{2}$, so $a = \sqrt{3}$.

Also, $\sin \alpha = \sin \left(-\frac{\pi}{6}\right) = -\frac{1}{2} = \frac{b}{R} = \frac{b}{2}$, so $b = -1$.

Let's check if these values satisfy $\sqrt{3}a - b = 4$. We have $\sqrt{3}(\sqrt{3}) - (-1) = 3 + 1 = 4$, which is correct. Also $a^2 + b^2 = (\sqrt{3})^2 + (-1)^2 = 3 + 1 = 4$, which is correct.

(2) We have $y = \sqrt{3} \sin \theta - \cos \theta$. Since $y = 2 \sin (\theta - \frac{\pi}{6})$, the minimum value of $y$ is $-2$, which occurs when $\sin (\theta - \frac{\pi}{6}) = -1$.

This means $\theta - \frac{\pi}{6} = \frac{3\pi}{2} + 2n\pi$ for some integer $n$. Thus, $\theta = \frac{3\pi}{2} + \frac{\pi}{6} + 2n\pi = \frac{9\pi + \pi}{6} + 2n\pi = \frac{10\pi}{6} + 2n\pi = \frac{5\pi}{3} + 2n\pi$.

Since $0 \le \theta < 2\pi$, we can take $n=0$ to get $\theta = \frac{5\pi}{3}$.

3. Final Answer

(1) $a = \sqrt{3}$, $b = -1$

(2) The minimum value of $y$ is $-2$, which occurs at $\theta = \frac{5\pi}{3}$.