3.113. $\frac{1+\cos \alpha+\cos 2 \alpha+\cos 3 \alpha}{\cos \alpha+2 \cos ^{2} \alpha-1}$.
## Solution.

$$
\begin{aligned}
& \frac{1+\cos \alpha+\cos 2 \alpha+\cos 3 \alpha}{\cos \alpha+2 \cos ^{2} \alpha-1}=\frac{1+\cos 2 \alpha+(\cos \alpha+\cos 3 \alpha)}{\cos \alpha+2 \cos ^{2} \alpha-1}= \\
& =\frac{1+\cos 2 \alpha+2 \cos 2 \alpha \cos \alpha}{\cos \alpha+2 \cos ^{2} \alpha-1}=\frac{1+2 \cos ^{2} \alpha-1+2\left(2 \cos ^{2} \alpha-1\right) \cos \alpha}{\cos \alpha+2 \cos ^{2} \alpha-1}= \\
& =\frac{2 \cos \alpha\left(\cos \alpha+2 \cos ^{2} \alpha-1\right)}{\cos \alpha+2 \cos ^{2} \alpha-1}=2 \cos \alpha .
\end{aligned}
$$

Answer: $2 \cos \alpha$.

Transform into a product (3.114-3.147):