Simplify the following expressions to their simplest form:

a) $\frac{\cos 3 \alpha+\cos \alpha}{2 \cos ^{3} \alpha-\sin \alpha \sin 2 \alpha}$,

b) $\frac{\sin \alpha \sin (\alpha+\beta)+\cos \alpha \cos (\alpha+\beta)}{\cos \beta}$,

c) $\frac{\sin 3 \alpha+4 \sin ^{3} \alpha}{\cos 3 \alpha-4 \cos ^{3} \alpha}+\operatorname{tg} \alpha$.
a) By transforming the numerator into a product using the formula $\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}$:

$$
\frac{2 \cos \frac{3 \alpha+\alpha}{2} \cos \frac{3 \alpha-\alpha}{2}}{2 \cos ^{3} \alpha-2 \sin ^{2} \alpha \cos \alpha}=\frac{2 \cos 2 \alpha \cos \alpha}{2 \cos \alpha\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)}=\frac{2 \cos 2 \alpha \cos \alpha}{2 \cos \alpha \cos 2 \alpha}=1
$$

b) It is known that $\cos (x-y)=\cos x \cos y+\sin x \sin y$, so in reverse, $\cos (\alpha+\beta) \cos \alpha+\sin (\alpha+\beta) \sin \alpha=\cos [(\alpha+\beta)-\alpha]=\cos \beta$, and thus our expression is

$$
\frac{\cos \beta}{\cos \beta}=1
$$

c)

$$
\begin{gathered}
\sin 3 \alpha=\sin (2 \alpha+\alpha)=\sin 2 \alpha \cos \alpha+\cos 2 \alpha \sin \alpha=2 \sin \alpha \cos ^{2} \alpha+\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right) \sin \alpha= \\
=2 \sin \alpha\left(1-\sin ^{2} \alpha\right)+\left(1-2 \sin ^{2} \alpha\right) \sin \alpha= \\
=2 \sin \alpha-2 \sin ^{3} \alpha+\sin \alpha-2 \sin ^{3} \alpha=3 \sin \alpha-4 \sin ^{3} \alpha \\
\cos 3 \alpha=\cos (2 \alpha+\alpha)=\cos 2 \alpha \cos \alpha-\sin 2 \alpha \sin \alpha=\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right) \cos \alpha-2 \sin ^{2} \alpha \cos \alpha= \\
=\cos ^{3} \alpha-\left(1-\cos ^{2} \alpha\right) \cos \alpha-2\left(1-\cos ^{2} \alpha\right) \cos \alpha= \\
=\cos ^{3} \alpha-\cos \alpha+\cos ^{3} \alpha-2 \cos \alpha+2 \cos ^{3} \alpha=4 \cos ^{3} \alpha-3 \cos \alpha
\end{gathered}
$$

Substituting the values we obtained into our expression:

$$
\frac{3 \sin \alpha-4 \sin ^{3} \alpha+4 \sin ^{3} \alpha}{4 \cos ^{3} \alpha-3 \cos \alpha-4 \cos ^{3} \alpha}+\operatorname{tg} \alpha=\frac{3 \sin \alpha}{-3 \cos \alpha}+\operatorname{tg} \alpha=-\operatorname{tg} \alpha+\operatorname{tg} \alpha=0
$$

Béla Rozsondai (Sopron, Széchenyi g. III. o. t.)