Question

Hi! I was absent today and did not understand this lesson please I will be really grateful if you help me ! I appreciate it this is classwork assignment does not count as a test

Answer 1

Given:

`\begin{gathered} \sin \alpha=(40)/(41)first\text{ quadrant} \\ \sin \beta=(4)/(5),\sec ondquadrant \end{gathered}`

Step 1:

Figure out the value of cos alpha

We will use the Pythagoras theorem below

`\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}`

Hence,

`\begin{gathered} \cos \alpha=\frac{\text{adjacent}}{\text{hypotenus}} \\ \cos \alpha=(9)/(41) \end{gathered}`

Step 2:

Figure out the value of cos beta

To figure this out, we will use the Pythagoras theorem below

`\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}`

Hence,

`\begin{gathered} \cos \beta=\frac{\text{adjacent}}{\text{hypotenus}} \\ \cos \beta=-(3)/(5)(\cos \text{ is negative on the second quadrant)} \end{gathered}`

Step 3:

`\cos (\alpha+\beta)=\cos \alpha\cos \beta-\sin \alpha\sin \beta`

By substituting the values, we will have

`\begin{gathered} \cos (\alpha+\beta)=\cos \alpha\cos \beta-\sin \alpha\sin \beta \\ \cos (\alpha+\beta)=(9)/(41)*-(3)/(5)-(40)/(41)*(4)/(5) \\ \cos (\alpha+\beta)=-(27)/(205)-(160)/(205) \\ \cos (\alpha+\beta)=-(187)/(205) \end{gathered}`

Hence,

The final answer = -187/205