Answer:
[tex]\displaystyle \cos(\alpha+\beta)=\frac{3\sqrt{22}-\sqrt{371}}{40}[/tex]
Step-by-step explanation:
We are given that:
[tex]\displaystyle \cos(\alpha)=\frac{\sqrt{11}}{8}\text{ and } \sin(\beta)=\frac{\sqrt7}{5}[/tex]
Where both α and β are in QI.
And we want to find cos(α + β).
First, let's determine the side lengths for each angle.
For α, we are given that its cosine is √(11)/8.
And since cosine is the ratio of the adjacent side to the hypotenuse, the adjacent side to α is √11 and the hypotenuse is 8.
Therefore, the opposite side will be:
[tex]o=\sqrt{8^2-(\sqrt{11})^2}=\sqrt{53}[/tex]
Hence, for α, the adjacent side is √11, the opposite side is √53, and the hypotenuse is 8.
Likewise, for β, we are given that its sine is √7/5.
And since sine is the ratio of the opposite side to the hypotenuse, the adjacent side of β is:
[tex]a=\sqrt{5^2-(\sqrt{7})^2}=\sqrt{18}=3\sqrt{2}[/tex]
In summary:
For α, the adjacent is √11, the opposite is √53, and the hypotenuse is 8.
For β, the adjacent is 3√2, the opposite is √7, and the hypotenuse is 5.
Using an angle addition identity, we can rewrite our expression as:
[tex]\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)[/tex]
And since both α and β are in QI, all trig ratios will be positive.
Using the above information, we can substitute in the following values:
[tex]\displaystyle \cos(\alpha +\beta)=\Big(\frac{\sqrt{11}}{8}\Big)\Big(\frac{3\sqrt2}{5}\Big)-\Big(\frac{\sqrt{53}}{8}\Big)\Big(\frac{\sqrt7}{5}\Big)[/tex]
Finally, simplify:
[tex]\displaystyle \cos(\alpha +\beta)=\frac{3\sqrt{22}}{40}-\frac{\sqrt{371}}{40}=\frac{3\sqrt{22}-\sqrt{371}}{40}\approx -0.1298[/tex]
