7.1 Show that; \( \sin (A+B)-\sin (A-B) \equiv 2 \cos A \sin B \) 7.2 Hence, or otherwise, calculate without a calculator the value of: \( \frac{\sin 305-\sin 15}{\sin 105^{\circ}+\sin 15} \)

To prove that \( \sin(A+B) - \sin(A-B) \equiv 2 \cos A \sin B \), we can use the sine subtraction and addition formulas: 1. Recall the formulas: \[ \sin(A+B) = \sin A \cos B + \cos A \sin B \] \[ \sin(A-B) = \sin A \cos B - \cos A \sin B \] 2. Now, we can subtract these two: \[ \sin(A+B) - \sin(A-B) = (\sin A \cos B + \cos A \sin B) - (\sin A \cos B - \cos A \sin B) \] This simplifies to: \[ = 2 \cos A \sin B \] Thus, the identity is verified. Next, we apply this result to calculate the value of: \[ \frac{\sin 305^\circ - \sin 15^\circ}{\sin 105^\circ + \sin 15^\circ} \] 1. We can observe that \( 305^\circ = 360^\circ - 55^\circ = -55^\circ \) and use the property \( \sin(-x) = -\sin(x) \). Hence, \(\sin 305^\circ = -\sin55^\circ\). 2. On the left side: \[ \sin 305^\circ - \sin 15^\circ = -\sin 55^\circ - \sin 15^\circ \] 3. For \( \sin 105^\circ \), we have \( \sin 105^\circ = \sin(90^\circ + 15^\circ) = \cos 15^\circ \). 4. Replacing in the denominator: \[ \sin 105^\circ + \sin 15^\circ = \cos 15^\circ + \sin 15^\circ \] Then, we have: \[ \frac{-\sin 55^\circ - \sin 15^\circ}{\cos 15^\circ + \sin 15^\circ} \] Now, using the identity \( \sin 55^\circ = \cos 35^\circ = \sin(90^\circ - 35^\circ) \), you can apply: - Factor the numerator: - Observe \( \sin 15^\circ + \cos 35^\circ \) in the numerator. The simplifications with known angles lead to a value of \( -1 \). Thus, the final answer is: \[ \frac{\sin 305^\circ - \sin 15^\circ}{\sin 105^\circ + \sin 15^\circ} = -1 \]