Since
[tex]\begin{gathered} \sin (\alpha)=\frac{40}{41}, \\ \sin (\beta)=\frac{15}{17}, \end{gathered}[/tex]then
[tex]\begin{gathered} \cos (\alpha)=\frac{\sqrt[]{41^2-40^2}}{41}=\frac{9}{41}, \\ \cos (\beta)=\frac{\sqrt[]{17^2-15^2}}{17}=\frac{8}{17}\text{.} \end{gathered}[/tex]Now, recall that:
[tex]\begin{gathered} \cos (\alpha+\beta)=\cos (\alpha)\cos (\beta)-\sin (\alpha)\sin (\beta), \\ \sin (\alpha+\beta)=\sin (\alpha)\cos (\beta)+\cos (\alpha)\sin (\beta), \\ \tan (\alpha+\beta)=\frac{\tan (\alpha)+\tan (\beta)}{1-\text{tan(}\alpha)\tan (\beta)}\text{.} \end{gathered}[/tex]Therefore:
[tex]\cos (\alpha+\beta)=\frac{9}{41}\times\frac{8}{17}-\frac{40}{41}\times\frac{15}{17}=\frac{72}{697}-\frac{600}{697}=-\frac{528}{697}\text{.}[/tex]Answer:
[tex]\cos (\alpha+\beta)=-\frac{528}{697}\text{.}[/tex]