The problem requires us to rewrite the given trigonometric expressions in the form $r \sin(x+\alpha)$, where $r > 0$ and $-\pi < \alpha \le \pi$. The two expressions are: (1) $\sin x + \cos x$ (2) $3\sin x - \sqrt{3} \cos x$

1. Problem Description

The problem requires us to rewrite the given trigonometric expressions in the form $r \sin(x+\alpha)$, where $r > 0$ and $-\pi < \alpha \le \pi$. The two expressions are:

(1) $\sin x + \cos x$

(2) $3\sin x - \sqrt{3} \cos x$

2. Solution Steps

(1) $\sin x + \cos x$

We want to express this in the form $r\sin(x+\alpha) = r(\sin x \cos \alpha + \cos x \sin \alpha) = (r\cos \alpha)\sin x + (r\sin \alpha)\cos x$.

Comparing coefficients, we have

$r\cos \alpha = 1$

$r\sin \alpha = 1$

Squaring and adding the two equations, we get

$r^2\cos^2 \alpha + r^2\sin^2 \alpha = 1^2 + 1^2$

$r^2(\cos^2 \alpha + \sin^2 \alpha) = 2$

$r^2 = 2$

Since $r>0$, we have $r = \sqrt{2}$.

Then $\cos \alpha = \frac{1}{\sqrt{2}}$ and $\sin \alpha = \frac{1}{\sqrt{2}}$.

Thus, $\alpha = \frac{\pi}{4}$, which satisfies $-\pi < \alpha \le \pi$.

Therefore, $\sin x + \cos x = \sqrt{2}\sin(x + \frac{\pi}{4})$.

(2) $3\sin x - \sqrt{3} \cos x$

We want to express this in the form $r\sin(x+\alpha) = r(\sin x \cos \alpha + \cos x \sin \alpha) = (r\cos \alpha)\sin x + (r\sin \alpha)\cos x$.

Comparing coefficients, we have

$r\cos \alpha = 3$

$r\sin \alpha = -\sqrt{3}$

Squaring and adding the two equations, we get

$r^2\cos^2 \alpha + r^2\sin^2 \alpha = 3^2 + (-\sqrt{3})^2$

$r^2(\cos^2 \alpha + \sin^2 \alpha) = 9 + 3$

$r^2 = 12$

Since $r>0$, we have $r = \sqrt{12} = 2\sqrt{3}$.

Then $\cos \alpha = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$ and $\sin \alpha = \frac{-\sqrt{3}}{2\sqrt{3}} = -\frac{1}{2}$.

Thus, $\alpha = -\frac{\pi}{6}$, which satisfies $-\pi < \alpha \le \pi$.

Therefore, $3\sin x - \sqrt{3} \cos x = 2\sqrt{3}\sin(x - \frac{\pi}{6})$.

3. Final Answer

(1) $\sqrt{2}\sin(x + \frac{\pi}{4})$

(2) $2\sqrt{3}\sin(x - \frac{\pi}{6})$