Simplify the following trigonometric expressions: 7.1 ( 30^)/( 60^) + (360^ - theta) * 2theta. 7.2.1 Expand using the compound angle identity for sine.

We know that $\sin 30^\circ = \frac{1}{2}$ and $\cos 60^\circ = \frac{1}{2}$.
Also, $\tan(360^\circ - \theta) = -\tan \theta$ (tangent is negative in the fourth quadrant).
The expression becomes:

$$\frac{\sin 30^\circ}{\cos 60^\circ} + \tan(360^\circ - \theta) \cdot \sin 2\theta$$ $$= \frac{\frac{1}{2}}{\frac{1}{2}} + (-\tan \theta) \cdot \sin 2\theta$$ $$= 1 - \tan \theta \cdot \sin 2\theta$$

The identity for $\sin 2\theta$ is $2 \sin \theta \cos \theta$.
The identity for $\tan \theta$ is $\frac{\sin \theta}{\cos \theta}$.
Substitute these into the expression:

$$1 - \left(\frac{\sin \theta}{\cos \theta}\right) \cdot (2 \sin \theta \cos \theta)$$

The $\cos \theta$ terms cancel out:

$$1 - 2 \sin \theta \sin \theta$$ $$= 1 - 2 \sin^2 \theta$$

The expression $1 - 2 \sin^2 \theta$ is a double angle identity for $\cos 2\theta$.

$$\cos 2\theta = 1 - 2 \sin^2 \theta$$

Thus, the simplified expression is:

$$\boxed{\cos 2\theta}$$

The compound angle identity is $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$.
For $\sin(60^\circ + 2x)$:

$$\sin(60^\circ + 2x) = \sin 60^\circ \cos 2x + \cos 60^\circ \sin 2x$$

For $\sin(60^\circ - 2x)$:

$$\sin(60^\circ - 2x) = \sin 60^\circ \cos 2x - \cos 60^\circ \sin 2x$$

$$\sin(60^\circ + 2x) + \sin(60^\circ - 2x) = (\sin 60^\circ \cos 2x + \cos 60^\circ \sin 2x) + (\sin 60^\circ \cos 2x - \cos 60^\circ \sin 2x)$$

The terms $\cos 60^\circ \sin 2x$ cancel out:

$$= 2 \sin 60^\circ \cos 2x$$

We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$.

$$= 2 \left(\frac{\sqrt{3}}{2}\right) \cos 2x$$ $$= \sqrt{3} \cos 2x$$

Given that $\sin(60^\circ + 2x) + \sin(60^\circ - 2x) = k \cos 2x$.
From our simplification, we have:

$$\sqrt{3} \cos 2x = k \cos 2x$$

By comparing the coefficients of $\cos 2x$ on both sides, we find:

$$k = \sqrt{3}$$

The value of $k$ is:

$$\boxed{\sqrt{3}}$$