Key Concepts and Formulas

• Trigonometric identities, specifically the sine of supplementary angles: $\sin(180^\circ - x) = \sin(x)$
• Sum of angles in a quadrilateral: $A + B + C + D = 360^\circ = 2\pi$
• Cosine of supplementary angles: $\cos(180^\circ - x) = -\cos(x)$

Step-by-Step Solution

Statement p:

• Step 1: Evaluate $\sin 120^\circ$. We know that $\sin 120^\circ = \sin (180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$.

• Step 2: Evaluate $2\sin 120^\circ$. Multiplying the result from Step 1 by 2, we get $2\sin 120^\circ = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$. This is the value we want to match with the right-hand side of the equation in statement p.

• Step 3: Evaluate $\sin 240^\circ$. We have $\sin 240^\circ = \sin (180^\circ + 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$.

• Step 4: Substitute $\theta = 240^\circ$ into the right-hand side of the equation and simplify. The expression becomes: $\sqrt{1 + \sin 240^\circ} - \sqrt{1 - \sin 240^\circ} = \sqrt{1 - \frac{\sqrt{3}}{2}} - \sqrt{1 + \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{2}} - \sqrt{\frac{2 + \sqrt{3}}{2}}$ Since $\sqrt{3} \approx 1.732$, we have $\frac{2 - \sqrt{3}}{2} > 0$ and $\frac{2 + \sqrt{3}}{2} > 0$. Also, $\sqrt{\frac{2 + \sqrt{3}}{2}} > \sqrt{\frac{2 - \sqrt{3}}{2}}$, so the entire expression is negative.

• Step 5: Compare the results. We calculated that $2\sin 120^\circ = \sqrt{3}$, which is positive. However, $\sqrt{1 + \sin 240^\circ} - \sqrt{1 - \sin 240^\circ}$ is negative. Therefore, the equation in statement p does not hold for $\theta = 240^\circ$.

• Step 6: Conclude on the truth value of p. Since the equation does not hold, statement p is false (F).

Statement q:

• Step 1: State the sum of angles in a quadrilateral. For any quadrilateral ABCD, the sum of its angles is $A + B + C + D = 360^\circ = 2\pi$.

• Step 2: Divide the equation by 2. Dividing both sides by 2, we get $\frac{A + B + C + D}{2} = \pi$, which can be rearranged as $\frac{A + C}{2} + \frac{B + D}{2} = \pi$.

• Step 3: Express $\frac{B + D}{2}$ in terms of $\frac{A + C}{2}$. From Step 2, we have $\frac{B + D}{2} = \pi - \frac{A + C}{2}$.

• Step 4: Substitute into the expression in statement q. We want to evaluate $\cos\left(\frac{A + C}{2}\right) + \cos\left(\frac{B + D}{2}\right)$. Substituting $\frac{B + D}{2} = \pi - \frac{A + C}{2}$, we get: $\cos\left(\frac{A + C}{2}\right) + \cos\left(\pi - \frac{A + C}{2}\right)$

• Step 5: Use the cosine supplementary angle identity. Since $\cos(\pi - x) = -\cos(x)$, we have: $\cos\left(\frac{A + C}{2}\right) + \cos\left(\pi - \frac{A + C}{2}\right) = \cos\left(\frac{A + C}{2}\right) - \cos\left(\frac{A + C}{2}\right) = 0$

• Step 6: Conclude on the truth value of q. Since the equation holds, statement q is true (T).

Common Mistakes & Tips

• Signs: Be very careful with the signs of trigonometric functions in different quadrants. For example, $\sin 240^\circ$ is negative.
• Angle Sum: Remember that the sum of angles in a quadrilateral is $360^\circ$ (or $2\pi$ radians).
• Supplementary Angles: The relationships between trigonometric functions of supplementary angles (angles that add up to $180^\circ$ or $\pi$ radians) are crucial.

Summary We evaluated statement p by calculating $\sin 120^\circ$ and $\sin 240^\circ$, then substituting into the given equation. We found that the equation did not hold, so statement p is false. For statement q, we used the fact that the sum of the angles in a quadrilateral is $360^\circ$ to show that $\cos\left(\frac{A + C}{2}\right) + \cos\left(\frac{B + D}{2}\right) = 0$, so statement q is true. Therefore, the truth values of p and q are F and T, respectively.

Final Answer The final answer is \boxed{F, T}, which corresponds to option (A).