[Algebra]
Problem: Compute the value of

$$
\sin \left(6^{\circ}\right) \cdot \sin \left(12^{\circ}\right) \cdot \sin \left(24^{\circ}\right) \cdot \sin \left(42^{\circ}\right)+\sin \left(12^{\circ}\right) \cdot \sin \left(24^{\circ}\right) \cdot \sin \left(42^{\circ}\right) \text {. }
$$

Solution: Let $S=\left(1+\sin 6^{\circ}\right)\left(\sin 12^{\circ} \sin 24^{\circ} \sin 42^{\circ}\right)$. It follows from a sum-to-product identity that $1+\sin 6^{\circ}=$ $\sin 90^{\circ}+\sin 6^{\circ}=2 \sin 48^{\circ} \cos 42^{\circ}$. Because the sine of an angle is the cosine of its complement, it follows that

$$
S=\left(2 \sin 48^{\circ} \cos 42^{\circ}\right)\left(\sin 12^{\circ} \sin 24^{\circ} \sin 42^{\circ}\right)=2\left(\sin 48^{\circ}\right)^{2}\left(\sin 12^{\circ} \sin 24^{\circ} \cos 48^{\circ}\right)
$$

By the double-angle formula, this means $S=\sin 12^{\circ} \sin 24^{\circ} \sin 48^{\circ} \sin 96^{\circ}$. By a product-to-sum identity,

$$
\sin 12^{\circ} \sin 48^{\circ}=\frac{\cos 36^{\circ}-\cos 60^{\circ}}{2}=\frac{\sqrt{5}-1}{8}
$$



and

$$
\sin 24^{\circ} \sin 96^{\circ}=\frac{\cos 72^{\circ}-\cos 120^{\circ}}{2}=\frac{\sqrt{5}+1}{8}
$$

Multiply the expressions on the right-hand sides of (1) and (2) to obtain $\frac{\mathbf{1}}{\mathbf{1 6}}$

Final answer: $\frac{1}{16}$