Given theta = -(5)/(13) where 180^ < theta < 360^, find the values of 2 theta and (theta - 135^).

Here are the solutions to the problems.

6.1 Given $\cos \theta = -\frac{5}{13}$ where $180^\circ < \theta < 360^\circ$.
Since $\cos \theta$ is negative and $180^\circ < \theta < 360^\circ$, $\theta$ must be in Quadrant III. In Quadrant III, $\sin \theta$ is negative.

6.1.1 Determine the value of $\sin^2 \theta$.

Step 1: Use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$.

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

Step 2: Substitute the given value of $\cos \theta$.

$$\sin^2 \theta = 1 - \left(-\frac{5}{13}\right)^2$$ $$\sin^2 \theta = 1 - \frac{25}{169}$$

Step 3: Simplify the expression.

$$\sin^2 \theta = \frac{169}{169} - \frac{25}{169}$$ $$\sin^2 \theta = \frac{144}{169}$$

The value of $\sin^2 \theta$ is $\boxed{\frac{144}{169}}$.

6.1.2 Determine the value of $\cos(\theta - 135^\circ)$.

Step 1: Find the value of $\sin \theta$. Since $\sin^2 \theta = \frac{144}{169}$ and $\theta$ is in Quadrant III, $\sin \theta$ must be negative.

$$\sin \theta = -\sqrt{\frac{144}{169}} = -\frac{12}{13}$$

Step 2: Use the compound angle formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$.
Here, $A = \theta$ and $B = 135^\circ$.
We need the values for $\cos 135^\circ$ and $\sin 135^\circ$.
$\cos 135^\circ = \cos(180^\circ - 45^\circ) = -\cos 45^\circ = -\frac{\sqrt{2}}{2}$
$\sin 135^\circ = \sin(180^\circ - 45^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2}$

Step 3: Substitute the values into the formula.

$$\cos(\theta - 135^\circ) = \left(-\frac{5}{13}\right)\left(-\frac{\sqrt{2}}{2}\right) + \left(-\frac{12}{13}\right)\left(\frac{\sqrt{2}}{2}\right)$$

Step 4: Simplify the expression.

$$\cos(\theta - 135^\circ) = \frac{5\sqrt{2}}{26} - \frac{12\sqrt{2}}{26}$$ $$\cos(\theta - 135^\circ) = \frac{5\sqrt{2} - 12\sqrt{2}}{26}$$ $$\cos(\theta - 135^\circ) = \frac{-7\sqrt{2}}{26}$$

The value of $\cos(\theta - 135^\circ)$ is $\boxed{-\frac{7\sqrt{2}}{26}}$.

6.2 Prove that $2\cos^2(45^\circ + x) = 1 - \sin 2x$.

Step 1: Start with the left-hand side (LHS) and use the double angle identity $2\cos^2 A = 1 + \cos 2A$, where $A = 45^\circ + x$.

$$\text{LHS} = 2\cos^2(45^\circ + x)$$ $$\text{LHS} = 1 + \cos(2(45^\circ + x))$$

Step 2: Simplify the argument of the cosine function.

$$\text{LHS} = 1 + \cos(90^\circ + 2x)$$

Step 3: Use the co-function identity $\cos(90^\circ + \alpha) = -\sin \alpha$, where $\alpha = 2x$.

$$\text{LHS} = 1 - \sin 2x$$

Step 4: The LHS is equal to the right-hand side (RHS).

$$\text{LHS} = \text{RHS}$$

Therefore, $2\cos^2(45^\circ + x) = 1 - \sin 2x$ is proven.

6.3 Consider the expression: $\sin(A - B) - \sin(A + B)$.

6.3.1 Prove that $\sin(A - B) - \sin(A + B) = -2\cos A \sin B$.

Step 1: Start with the left-hand side (LHS) and expand using the compound angle formulas.

$$\text{LHS} = \sin(A - B) - \sin(A + B)$$ $$\text{LHS} = (\sin A \cos B - \cos A \sin B) - (\sin A \cos B + \cos A \sin B)$$

Step 2: Distribute the negative sign and simplify.

$$\text{LHS} = \sin A \cos B - \cos A \sin B - \sin A \cos B - \cos A \sin B$$ $$\text{LHS} = (\sin A \cos B - \sin A \cos B) - (\cos A \sin B + \cos A \sin B)$$ $$\text{LHS} = 0 - 2\cos A \sin B$$ $$\text{LHS} = -2\cos A \sin B$$

Step 3: The LHS is equal to the right-hand side (RHS).

$$\text{LHS} = \text{RHS}$$

Therefore, $\sin(A - B) - \sin(A + B) = -2\cos A \sin B$ is proven.

6.3.2 Simplify the following expression to a single term: $\sin 4x - \sin 10x$.

Step 1: Use the sum-to-product formula $\sin P - \sin Q = 2\cos\left(\frac{P+Q}{2}\right)\sin\left(\frac{P-Q}{2}\right)$.
Here, $P = 4x$ and $Q = 10x$.

Step 2: Substitute $P$ and $Q$ into the formula.

$$\sin 4x - \sin 10x = 2\cos\left(\frac{4x+10x}{2}\right)\sin\left(\frac{4x-10x}{2}\right)$$

Step 3: Simplify the arguments.

$$\sin 4x - \sin 10x = 2\cos\left(\frac{14x}{2}\right)\sin\left(\frac{-6x}{2}\right)$$ $$\sin 4x - \sin 10x = 2\cos(7x)\sin(-3x)$$

Step 4: Use the identity $\sin(-\alpha) = -\sin \alpha$.

$$\sin 4x - \sin 10x = 2\cos(7x)(-\sin 3x)$$ $$\sin 4x - \sin 10x = -2\cos(7x)\sin(3x)$$

The simplified expression is $\boxed{-2\cos(7x)\sin(3x)}$.

6.3.3 Hence, determine the solution for $\sin 4x - \sin 10x = \sin 3x$ for $x \in [0^\circ; 30^\circ]$.

Step 1: Substitute the simplified expression from 6.3.2 into the equation.

$$-2\cos(7x)\sin(3x) = \sin(3x)$$

Step 2: Rearrange the equation to set it to zero and factor out the common term $\sin(3x)$.

$$-2\cos(7x)\sin(3x) - \sin(3x) = 0$$ $$\sin(3x)(-2\cos(7x) - 1) = 0$$

Step 3: Set each factor equal to zero and solve for $x$.
Case 1: $\sin(3x) = 0$
The general solution for $\sin \alpha = 0$ is $\alpha = n \cdot 180^\circ$, where $n$ is an integer.

$$3x = n \cdot 180^\circ$$ $$x = n \cdot 60^\circ$$

For $x \in [0^\circ; 30^\circ]$:
If $n=0$, $x = 0^\circ$.
If $n=1$, $x = 60^\circ$ (outside the interval).
So, $x = 0^\circ$ is a solution.

Case 2: $-2\cos(7x) - 1 = 0$

$$-2\cos(7x) = 1$$ $$\cos(7x) = -\frac{1}{2}$$

The reference angle for $\cos \alpha = \frac{1}{2}$ is $60^\circ$. Since $\cos(7x)$ is negative, $7x$ is in Quadrant II or III.
General solutions for $\cos \alpha = -\frac{1}{2}$:
$\alpha = 180^\circ - 60^\circ + n \cdot 360^\circ = 120^\circ + n \cdot 360^\circ$
$\alpha = 180^\circ + 60^\circ + n \cdot 360^\circ = 240^\circ + n \cdot 360^\circ$
So, for $7x$:

$$7x = 120^\circ + n \cdot 360^\circ \quad \text{or} \quad 7x = 240^\circ + n \cdot 360^\circ$$ $$$$