a) It is to be proven that

$$
\sin ^{2} 19^{\circ}+\sin 19^{\circ} \sin 41^{\circ}+\sin ^{2} 41^{\circ}=\frac{3}{4}
$$

b) What is the value of the following expression:

$$
\cos ^{2} 74^{\circ}+\cos 74^{\circ} \cos 46^{\circ}+\cos ^{2} 46^{\circ} ?
$$
a) Since $41^{\circ}=60^{\circ}-19^{\circ}$, therefore

$$
\begin{aligned}
\sin 41^{\circ} & =\frac{1}{2}\left(\sqrt{3} \cos 19^{\circ}-\sin 19^{\circ}\right)=\frac{1}{4}\left(2 \sqrt{3} \cos 19^{\circ}-2 \sin 19^{\circ}\right) \\
\sin ^{2} 41^{\circ} & =\frac{1}{4}\left(3 \cos ^{2} 19^{\circ}-2 \sqrt{3} \sin 19^{\circ} \cos 19^{\circ}+\sin ^{2} 19^{\circ}\right)
\end{aligned}
$$

It can be seen that by substituting these into the left side of (1), the sum of the terms containing the factor $\sqrt{3}$ is 0, and thus

$$
\frac{3}{4}\left(\sin ^{2} 19^{\circ}+\cos ^{2} 19^{\circ}\right)=\frac{3}{4}
$$

which is what we needed to prove.

The value of the $19^{\circ}$ angle was not used, so the following identity holds for any $x$:

$$
\sin ^{2} x+\sin x \sin \left(60^{\circ}-x\right)+\sin ^{2}\left(60^{\circ}-x\right)=\frac{3}{4}
$$

b) We can reduce the calculation of the value of $B$ in expression (2) to this:

$$
B=\sin ^{2} 16^{\circ}+\sin 16^{\circ} \sin 44^{\circ}+\sin ^{2} 44^{\circ}=\frac{3}{4}
$$