Mixed Exercise: (answel 1) Simplify the followi a) \( \frac{\cos \left(90^{\circ}+x\right)}{\sin ( } \) b) \( \cos 72^{\circ} \sin 48^{\circ} \)

**Part (a):** The expression provided appears incomplete. The numerator is \[ \cos\left(90^\circ+x\right) \] but the denominator is shown as \[ \sin( \] with no angle specified. Could you please confirm the complete expression for part (a)? **Part (b):** We want to simplify \[ \cos 72^\circ \sin 48^\circ. \] **Step 1. Use the product-to‐sum identity** Recall the product-to‐sum identity for the product of a cosine and a sine: \[ \cos A \sin B = \frac{1}{2}\Bigl[\sin(A+B) + \sin(B-A)\Bigr]. \] Here, let \[ A=72^\circ \quad \text{and} \quad B=48^\circ. \] **Step 2. Substitute the values of \(A\) and \(B\):** Substitute into the identity: \[ \cos 72^\circ \sin 48^\circ = \frac{1}{2}\Bigl[\sin(72^\circ+48^\circ) + \sin(48^\circ-72^\circ)\Bigr]. \] **Step 3. Simplify the angles:** Calculate the sums and differences: \[ 72^\circ+48^\circ = 120^\circ, \] \[ 48^\circ-72^\circ = -24^\circ. \] So, \[ \cos 72^\circ \sin 48^\circ = \frac{1}{2}\Bigl[\sin 120^\circ + \sin(-24^\circ)\Bigr]. \] **Step 4. Use known sine values and properties:** We know that \[ \sin 120^\circ = \sin(180^\circ-60^\circ) = \sin 60^\circ = \frac{\sqrt3}{2}, \] and \[ \sin(-24^\circ) = -\sin 24^\circ. \] Thus, the expression becomes: \[ \cos 72^\circ \sin 48^\circ = \frac{1}{2}\left(\frac{\sqrt3}{2} - \sin 24^\circ\right). \] **Step 5. Simplify the fraction:** Multiply the fraction: \[ \cos 72^\circ \sin 48^\circ = \frac{\sqrt3}{4} - \frac{1}{2}\sin 24^\circ. \] **Final Answer for Part (b):** \[ \cos 72^\circ \sin 48^\circ = \frac{\sqrt3}{4} - \frac{1}{2}\sin 24^\circ. \]