Answer:
Given:
[tex]\begin{gathered} \sin \alpha=\frac{40}{41}first\text{ quadrant} \\ \sin \beta=\frac{4}{5},\sec ondquadrant \end{gathered}[/tex]Step 1:
Figure out the value of cos alpha
We will use the Pythagoras theorem below
[tex]\begin{gathered} \text{hyp}^2=\text{opp}^2+\text{adj}^2 \\ \text{hyp}=41,\text{opp}=40,\text{adj}=x \\ 41^2=40^2+x^2 \\ 1681=1600+x^2 \\ x^2=1681-1600 \\ x^2=81 \\ x=\sqrt[]{81} \\ x=9 \end{gathered}[/tex]Hence,
[tex]\begin{gathered} \cos \alpha=\frac{\text{adjacent}}{\text{hypotenus}} \\ \cos \alpha=\frac{9}{41} \end{gathered}[/tex]Step 2:
Figure out the value of cos beta
To figure this out, we will use the Pythagoras theorem below
[tex]\begin{gathered} \text{hyp}^2=\text{opp}^2+\text{adj}^2 \\ \text{hyp}=5,\text{opp}=4,\text{adj}=y \\ 5^2=4^2+y^2 \\ 25=16+y^2 \\ y^2=25-16 \\ y^2=9 \\ y=\sqrt[]{9} \\ y=3 \end{gathered}[/tex]Hence,
[tex]\begin{gathered} \cos \beta=\frac{\text{adjacent}}{\text{hypotenus}} \\ \cos \beta=-\frac{3}{5}(\cos \text{ is negative on the second quadrant)} \end{gathered}[/tex]Step 3:
[tex]\cos (\alpha+\beta)=\cos \alpha\cos \beta-\sin \alpha\sin \beta[/tex]By substituting the values, we will have
[tex]\begin{gathered} \cos (\alpha+\beta)=\cos \alpha\cos \beta-\sin \alpha\sin \beta \\ \cos (\alpha+\beta)=\frac{9}{41}\times-\frac{3}{5}-\frac{40}{41}\times\frac{4}{5} \\ \cos (\alpha+\beta)=-\frac{27}{205}-\frac{160}{205} \\ \cos (\alpha+\beta)=-\frac{187}{205} \end{gathered}[/tex]Hence,
The final answer = -187/205
