We are asked to find $\cos(\alpha + \beta)$, given that:

$\cos(\alpha) = \frac{\sqrt{10}}{9}, \quad \sin(\beta) = \frac{\sqrt{10}}{9}$

Step 1: Use the cosine addition formula

The cosine addition formula is:

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

Step 2: Determine $\sin(\alpha)$ and $\cos(\beta)$

• Since $\alpha$ and $\beta$ are first quadrant angles, we know that both sine and cosine values are positive.

• We can find $\sin(\alpha)$ using the identity:

$\sin^2(\alpha) + \cos^2(\alpha) = 1$

Substitute $\cos(\alpha) = \frac{\sqrt{10}}{9}$:

$\sin^2(\alpha) + \left(\frac{\sqrt{10}}{9}\right)^2 = 1$

$\sin^2(\alpha) + \frac{10}{81} = 1$

$\sin^2(\alpha) = 1 - \frac{10}{81} = \frac{81}{81} - \frac{10}{81} = \frac{71}{81}$

$\sin(\alpha) = \sqrt{\frac{71}{81}} = \frac{\sqrt{71}}{9}$

• Now, find $\cos(\beta)$ using the same identity:

$\sin^2(\beta) + \cos^2(\beta) = 1$

Substitute $\sin(\beta) = \frac{\sqrt{10}}{9}$:

$\left(\frac{\sqrt{10}}{9}\right)^2 + \cos^2(\beta) = 1$

$\frac{10}{81} + \cos^2(\beta) = 1$

$\cos^2(\beta) = 1 - \frac{10}{81} = \frac{71}{81}$

$\cos(\beta) = \frac{\sqrt{71}}{9}$

Step 3: Apply the values into the formula

Now substitute $\cos(\alpha)$, $\sin(\alpha)$, $\cos(\beta)$, and $\sin(\beta)$ into the cosine addition formula:

$\cos(\alpha + \beta) = \left(\frac{\sqrt{10}}{9}\right)\left(\frac{\sqrt{71}}{9}\right) - \left(\frac{\sqrt{71}}{9}\right)\left(\frac{\sqrt{10}}{9}\right)$

Both terms are the same, so:

$\cos(\alpha + \beta) = 0$

Final Answer:

$\cos(\alpha + \beta) = 0$

Do you want more details or have any questions about this solution? Here are some related questions:

1. How do we derive the cosine addition formula?
2. Why are sine and cosine positive in the first quadrant?
3. How do we solve for the other trigonometric function if one is given?
4. What are other identities involving cosine and sine for sums and differences?
5. Can we find $\sin(\alpha + \beta)$ as well?

Tip: Always remember to use the Pythagorean identity to find the missing trigonometric function when given one.