Trigonometric Calculation for α + β

Given the values:

• $\sin \alpha = \frac{1}{\sqrt{5}}$
• $\cos \beta = \frac{3}{\sqrt{10}}$

We need to find the value of $\alpha + \beta$. First, let's find the corresponding cosine value for $\alpha$ and sine value for $\beta$, assuming both angles are acute (in the first quadrant).

Finding Cosine of α

Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$:

$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{1}{\sqrt{5}}\right)^2 = 1 - \frac{1}{5} = \frac{4}{5}$

$\cos \alpha = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$

Finding Sine of β

Using the identity $\sin^2 \beta + \cos^2 \beta = 1$:

$\sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(\frac{3}{\sqrt{10}}\right)^2 = 1 - \frac{9}{10} = \frac{1}{10}$

$\sin \beta = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$

Calculating α + β

Now, let's use the angle addition formula for cosine:

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

Substitute the known values:

$\cos(\alpha + \beta) = \left(\frac{2}{\sqrt{5}}\right) \left(\frac{3}{\sqrt{10}}\right) - \left(\frac{1}{\sqrt{5}}\right) \left(\frac{1}{\sqrt{10}}\right)$

$\cos(\alpha + \beta) = \frac{6}{\sqrt{50}} - \frac{1}{\sqrt{50}} = \frac{5}{\sqrt{50}}$

Simplify $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$:

$\cos(\alpha + \beta) = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$

The angle whose cosine is $\frac{1}{\sqrt{2}}$ is $45^{\circ}$.

Alternatively, using the angle addition formula for sine:

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

$\sin(\alpha + \beta) = \left(\frac{1}{\sqrt{5}}\right) \left(\frac{3}{\sqrt{10}}\right) + \left(\frac{2}{\sqrt{5}}\right) \left(\frac{1}{\sqrt{10}}\right)$

$\sin(\alpha + \beta) = \frac{3}{\sqrt{50}} + \frac{2}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$

Since both $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ are $\frac{1}{\sqrt{2}}$, the angle $\alpha + \beta$ is $45^{\circ}$.