142. Calculate, without using tables, the following products:

a) $A=\cos 4^{\circ} \cos 8^{\circ} \ldots \cos 84^{\circ} \cos 88^{\circ}$;

b) $B=\cos 12^{\circ} \cos 24^{\circ} \ldots \cos 72^{\circ} \cos 84^{\circ}$;

c) $C=\cos 1^{\circ} \cos 3^{\circ} \ldots \cos 87^{\circ} \cos 89^{\circ}$;

d) $D=\cos 5^{\circ} \cos 15^{\circ} \ldots \cos 87^{\circ} \cos 89^{\circ}$.
Solution. a) Multiply and divide the given expression by $2 \sin 4^{\circ}:$

$$
\begin{gathered}
A=\frac{2 \sin 4^{\circ} \cos 4^{\circ} \cos 8^{\circ} \ldots \cos 84^{\circ} \cos 88^{\circ}}{2 \sin 4^{\circ}}= \\
=\frac{2 \sin 8^{\circ} \cos 8^{\circ} \cos 16^{\circ} \ldots \cos 88^{\circ}}{2 \cdot 2 \sin 4^{\circ}}= \\
=\frac{2 \sin 16^{\circ} \cos 16^{\circ} \ldots \cos 88^{\circ}}{2 \cdot 2 \cdot 2 \sin 4^{\circ}}=\frac{\sin 176^{\circ}}{2^{22} \sin 4^{\circ}}=\left(\frac{1}{2}\right)^{22}
\end{gathered}
$$

b) Similarly

$$
B=\frac{2 \sin 12^{\circ} \cos 12^{\circ} \cos 24^{\circ} \ldots \cos 84^{\circ}}{2 \sin 12^{\circ}}=\frac{\sin 168^{\circ}}{2^{7} \sin 12^{\circ}}=\left(\frac{1}{2}\right)^{7}
$$

c) Multiply the equalities term by term:

$$
\begin{aligned}
& \cos 1^{\circ} \cos 89^{\circ}=\frac{1}{2} \cos 88^{\circ} \\
& \cos 3^{\circ} \cos 87^{\circ}=\frac{1}{2} \cos 84^{\circ} \\
& \cos 89^{\circ} \cos 1^{\circ}=\frac{1}{2} \cos 88^{\circ}
\end{aligned}
$$

We get:

$$
\begin{gathered}
C^{2}=\left(\cos 1^{\circ} \cos 3^{\circ} \ldots \cos 89^{\circ}\right)^{2}= \\
=\left(\frac{1}{2}\right)^{45} \cdot\left(\cos 88^{\circ} \cos 84^{\circ} \ldots \cos 4^{\circ}\right)^{2}=\left(\frac{1}{2}\right)^{45} \cdot A^{2}= \\
=\left(\frac{1}{2}\right)^{45} \cdot\left(\frac{1}{2}\right)^{44} ; \\
C=\left(\frac{1}{2}\right)^{45} \cdot \sqrt{2}
\end{gathered}
$$

d) We have:

$$
\begin{aligned}
\cos 5^{\circ} \cos 85^{\circ} & =\frac{1}{2} \cos 80^{\circ} \\
\cos 15^{\circ} \cos 75^{\circ} & =\frac{1}{2} \cos 60^{\circ}
\end{aligned}
$$

$$
\cos 85^{\circ} \cos 5^{\circ}=\frac{1}{2} \cos 80^{\circ}
$$

Therefore,

$D^{2}=\left(\cos 5^{\circ} \cos 15^{\circ} \ldots \cos 85^{\circ}\right)^{2}=\left(\frac{1}{2}\right)^{9} \cdot\left(\cos 80^{\circ} \cos 60^{\circ} \cos 40^{\circ} \cos 20^{\circ}\right)^{2}$. Since $\cos 80^{\circ} \cos 60^{\circ} \cos 40^{\circ} \cos 20^{\circ}=\left(\frac{1}{2}\right)^{4}$, then

$$
\begin{aligned}
& D^{2}=\left(\frac{1}{2}\right)^{9} \cdot\left(\frac{1}{2}\right)^{8}=2 \cdot\left(\frac{1}{2}\right)^{18}, D=\left(\frac{1}{2}\right)^{9} \sqrt{2} \\
& \text { Answer: a) }\left(\frac{1}{2}\right)^{22} ; \text { b) }\left(\frac{1}{2}\right)^{7} ; \text { c) }\left(\frac{1}{2}\right)^{45} \sqrt{2} ; \text { d) }\left(\frac{1}{2}\right)^{9} \sqrt{2}
\end{aligned}
$$