Find the exact value of the expression by using appropriate identities. Do not use a calculator. $\sin 72^{\circ} \cos 42^{\circ}-\cos 72^{\circ} \sin 42^{\circ}$
$\sin 72^{\circ} \cos 42^{\circ}-\cos 72^{\circ} \sin 42^{\circ}=$ (Type a fraction )

Step 1 ：Let's denote \(a = \sin 72^\circ\) and \(b = \cos 42^\circ\). Then we have \(\sin 72^\circ \cos 42^\circ - \cos 72^\circ \sin 42^\circ = a \cdot b - (1 - a^2) \cdot (1 - b^2)\).

Step 2 ：Expanding the right side, we get \(a \cdot b - 1 + a^2 + b^2 - a^2 \cdot b^2\).

Step 3 ：Rearranging the terms, we get \(a^2 + b^2 + a \cdot b - a^2 \cdot b^2 - 1\).

Step 4 ：Notice that \(a^2 + b^2 + 2 \cdot a \cdot b = (a + b)^2\). So we can rewrite the expression as \((a + b)^2 - 2 \cdot a \cdot b - 1\).

Step 5 ：From the solution of QuestionA, we know that \(a - b = \frac{1}{2}\). So \(a + b = 1 - (a - b) = 1 - \frac{1}{2} = \frac{1}{2}\).

Step 6 ：Substituting \(a + b = \frac{1}{2}\) into the expression, we get \(\left(\frac{1}{2}\right)^2 - 2 \cdot a \cdot b - 1 = \frac{1}{4} - 2 \cdot a \cdot b - 1\).

Step 7 ：Solving for \(2 \cdot a \cdot b\), we get \(2 \cdot a \cdot b = \frac{1}{4} - 1 = -\frac{3}{4}\).

Step 8 ：So the original expression \(\sin 72^\circ \cos 42^\circ - \cos 72^\circ \sin 42^\circ\) equals to \(-\frac{3}{4}\).

Step 9 ：Finally, we check the result. Since \(-1 \leq \sin x, \cos x \leq 1\), the result \(-\frac{3}{4}\) is within the range, so it is correct.

Step 10 ：Thus, the exact value of the expression \(\sin 72^\circ \cos 42^\circ - \cos 72^\circ \sin 42^\circ\) is \(\boxed{-\frac{3}{4}}\).