QUESTION 1 1.1 If \( \sin 24^{\circ}=p \), express the following in terms of \( p \), without the use of a calculator \( 1.1 .1 \cos 24^{\circ} \) 1.1.2 \( \sin 12^{\circ} \cos 12^{\circ}-\sin \left(-66^{\circ}\right) \tan 204^{\circ} \) QUESTION 2

**1.1.1** We are given \[ \sin 24^\circ = p. \] Using the Pythagorean identity for sine and cosine, we have \[ \sin^2 24^\circ + \cos^2 24^\circ = 1. \] Substituting \( \sin 24^\circ = p \) gives \[ p^2 + \cos^2 24^\circ = 1, \] so \[ \cos^2 24^\circ = 1 - p^2. \] Since \( 24^\circ \) is an acute angle, \( \cos 24^\circ \) is positive. Therefore, \[ \cos 24^\circ = \sqrt{1-p^2}. \] --- **1.1.2** We need to simplify \[ \sin 12^\circ \cos 12^\circ - \sin(-66^\circ) \tan 204^\circ. \] 1. **Simplify \(\sin 12^\circ \cos 12^\circ\):** Recall the double-angle identity: \[ \sin 12^\circ \cos 12^\circ = \frac{1}{2}\sin(2 \times 12^\circ) = \frac{1}{2}\sin 24^\circ. \] Since \( \sin 24^\circ = p \), we have \[ \sin 12^\circ \cos 12^\circ = \frac{1}{2}p. \] 2. **Simplify \(-\sin(-66^\circ) \tan 204^\circ\):** a. Use the property of sine: \[ \sin(-66^\circ) = -\sin 66^\circ. \] b. Notice that \( \tan 204^\circ \) can be reduced using the periodicity of the tangent function (period \(180^\circ\)): \[ \tan 204^\circ = \tan (204^\circ - 180^\circ) = \tan 24^\circ. \] c. Thus, \[ -\sin(-66^\circ) \tan 204^\circ = -(-\sin 66^\circ) \tan 24^\circ = \sin 66^\circ \tan 24^\circ. \] 3. **Express \(\sin 66^\circ \tan 24^\circ\) in terms of \( p \):** a. Recognize that \[ \sin 66^\circ = \sin(90^\circ - 24^\circ) = \cos 24^\circ. \] b. And, by definition, \[ \tan 24^\circ = \frac{\sin 24^\circ}{\cos 24^\circ} = \frac{p}{\cos 24^\circ}. \] c. Therefore, \[ \sin 66^\circ \tan 24^\circ = \cos 24^\circ \cdot \frac{p}{\cos 24^\circ} = p. \] 4. **Combine the results:** The original expression becomes \[ \frac{1}{2}p + p = \frac{3}{2}p. \] --- **QUESTION 2** No specific question was provided for Question 2. Please provide the details or statement of Question 2.