2 Prove that: \( \quad \sin \left(45^{\circ}+x\right) \cdot \sin \left(45^{\circ}-x\right)=\frac{\cos 2 x}{2} \)

To prove the identity \( \sin(45^{\circ}+x) \cdot \sin(45^{\circ}-x) = \frac{\cos 2x}{2} \), let's use the sine addition and subtraction formulas. We know that: \[ \sin(a + b) = \sin a \cos b + \cos a \sin b \] \[ \sin(a - b) = \sin a \cos b - \cos a \sin b \] Taking \( a = 45^{\circ} \) and \( b = x \), we have: \[ \sin(45^{\circ}+x) = \sin 45^{\circ} \cos x + \cos 45^{\circ} \sin x = \frac{\sqrt{2}}{2} \cos x + \frac{\sqrt{2}}{2} \sin x = \frac{\sqrt{2}}{2} (\cos x + \sin x) \] Similarly, \[ \sin(45^{\circ}-x) = \sin 45^{\circ} \cos x - \cos 45^{\circ} \sin x = \frac{\sqrt{2}}{2} \cos x - \frac{\sqrt{2}}{2} \sin x = \frac{\sqrt{2}}{2} (\cos x - \sin x) \] Now, we multiply these two results: \[ \sin(45^{\circ}+x) \cdot \sin(45^{\circ}-x) = \left(\frac{\sqrt{2}}{2}(\cos x + \sin x)\right) \left(\frac{\sqrt{2}}{2}(\cos x - \sin x)\right) \] This simplifies to: \[ \frac{2}{4} \left((\cos x + \sin x)(\cos x - \sin x)\right) = \frac{1}{2} (\cos^2 x - \sin^2 x) \] Using the identity \( \cos^2 x - \sin^2 x = \cos 2x \): \[ \frac{1}{2} (\cos^2 x - \sin^2 x) = \frac{\cos 2x}{2} \] Thus, we have proven that: \[ \sin(45^{\circ}+x) \cdot \sin(45^{\circ}-x) = \frac{\cos 2x}{2} \] This concludes the proof!