Question

Cos ( α ) = 11/8 and sin ( β ) = √ 7/5 . Find cos ( α + β )

Answer 1

Answer:

$\displaystyle \cos(\alpha+\beta)=\frac{3\sqrt{22}-\sqrt{371}}{40}$

Step-by-step explanation:

We are given that:

$\displaystyle \cos(\alpha)=\frac{\sqrt{11}}{8}\text{ and } \sin(\beta)=\frac{\sqrt7}{5}$

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:

$o=\sqrt{8^2-(\sqrt{11})^2}=\sqrt{53}$

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:

$a=\sqrt{5^2-(\sqrt{7})^2}=\sqrt{18}=3\sqrt{2}$

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:

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

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:

$\displaystyle \cos(\alpha +\beta)=\Big(\frac{\sqrt{11}}{8}\Big)\Big(\frac{3\sqrt2}{5}\Big)-\Big(\frac{\sqrt{53}}{8}\Big)\Big(\frac{\sqrt7}{5}\Big)$

Finally, simplify:

$\displaystyle \cos(\alpha +\beta)=\frac{3\sqrt{22}}{40}-\frac{\sqrt{371}}{40}=\frac{3\sqrt{22}-\sqrt{371}}{40}\approx -0.1298$