5.1 If \( \sin 40^{\circ} \cdot \cos 22^{\circ}+\cos 40^{\circ} \cdot \sin 22^{\circ}=k \), determine without the use of a calculato value of the following in terms of \( k \). \( 5.1 .3 \quad \sin 14^{\circ} \cdot \cos 14^{\circ} \)

To solve for \( \sin 14^{\circ} \cdot \cos 14^{\circ} \) in terms of \( k \), we can utilize the double angle identity: \[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta). \] Setting \( \theta = 14^{\circ} \), we have: \[ \sin(2 \cdot 14^{\circ}) = \sin(28^{\circ}) = 2 \sin(14^{\circ}) \cos(14^{\circ). \] Thus, we need to express \( \sin(28^{\circ}) \) in a relatable format. Using the sine addition formula, we recognize: \[ \sin(40^{\circ} - 12^{\circ}) = \sin(40^{\circ}) \cos(12^{\circ}) - \cos(40^{\circ}) \sin(12^{\circ}). \] Now utilizing the given information \( k = \sin 40^{\circ} \cdot \cos 22^{\circ} + \cos 40^{\circ} \cdot \sin 22^{\circ} \), we know from the sine addition formula that: \[ k = \sin(40^{\circ} + 22^{\circ}) = \sin(62^{\circ}). \] After determining \( \sin(28^{\circ}) \), we can express it in terms of \( k \) through some manipulation and known relationships: To summarize, to find \( \sin(14^{\circ}) \cdot \cos(14^{\circ}) \): \[ \sin(14^{\circ}) \cdot \cos(14^{\circ}) = \frac{1}{2} \sin(28^{\circ}) = \frac{1}{2} \cdot 2k = k. \] Therefore, we conclude: \[ \sin 14^{\circ} \cdot \cos 14^{\circ} = \frac{k}{2}. \]