B2. Since $\alpha$ is an acute angle, $\cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{12}{13}$. Then we calculate $\sin (2 \alpha+\pi)=-\sin 2 \alpha=$ $-2 \sin \alpha \cos \alpha=-2 \cdot \frac{5}{13} \cdot \frac{12}{13}=-\frac{120}{169}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-15.jpg?height=63&width=1639&top_left_y=2010&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-15.jpg?height=51&width=1637&top_left_y=2065&top_left_x=278)

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-15.jpg?height=60&width=1642&top_left_y=2109&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-15.jpg?height=52&width=1639&top_left_y=2167&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-15.jpg?height=54&width=1639&top_left_y=2212&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-15.jpg?height=68&width=1641&top_left_y=2259&top_left_x=276)
B2. Calculate $\sin (2 \alpha+\pi)$, if $\sin \alpha=\frac{5}{13}$ and $\alpha$ is an acute angle.