Find the e xact value of cos75^circ cdot sin15^circ

To find the exact value of \( \cos 75^\circ \cdot \sin 15^\circ \), we can use trigonometric identities and properties. Specifically, we will use the product-to-sum identities and the angle subtraction formulas.<br /><br />First, recall the product-to-sum identity for cosine and sine:<br />\[<br />\cos A \cdot \sin B = \frac{1}{2} [\sin (A + B) - \sin (A - B)]<br />\]<br /><br />Here, \( A = 75^\circ \) and \( B = 15^\circ \). Applying the identity, we get:<br />\[<br />\cos 75^\circ \cdot \sin 15^\circ = \frac{1}{2} [\sin (75^\circ + 15^\circ) - \sin (75^\circ - 15^\circ)]<br />\]<br /><br />Simplify the arguments of the sine functions:<br />\[<br />\sin (75^\circ + 15^\circ) = \sin 90^\circ<br />\]<br />\[<br />\sin (75^\circ - 15^\circ) = \sin 60^\circ<br />\]<br /><br />We know that:<br />\[<br />\sin 90^\circ = 1<br />\]<br />\[<br />\sin 60^\circ = \frac{\sqrt{3}}{2}<br />\]<br /><br />Substituting these values back into the expression, we have:<br />\[<br />\cos 75^\circ \cdot \sin 15^\circ = \frac{1}{2} [1 - \frac{\sqrt{3}}{2}]<br />\]<br /><br />Simplify the expression inside the brackets:<br />\[<br />1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}<br />\]<br /><br />Now, multiply by \(\frac{1}{2}\):<br />\[<br />\cos 75^\circ \cdot \sin 15^\circ = \frac{1}{2} \cdot \frac{2 - \sqrt{3}}{2} = \frac{2 - \sqrt{3}}{4}<br />\]<br /><br />Thus, the exact value of \( \cos 75^\circ \cdot \sin 15^\circ \) is:<br />\[<br />\boxed{\frac{2 - \sqrt{3}}{4}}<br />\]