## Task B-2.1.

If $\alpha$ and $\beta$ are acute angles of a right triangle and $\sin \beta=\frac{\sqrt{6}+\sqrt{2}}{4}$, determine $\sin \alpha$ and $\cos \alpha$.
## First Solution.

From the basic identity $\sin ^{2} \alpha+\cos ^{2} \alpha=1$ and $\cos \alpha=\sin \beta$, it follows that

$$
\sin ^{2} \alpha=1-\cos ^{2} \alpha=1-\sin ^{2} \beta
$$

Furthermore,

$$
\sin ^{2} \alpha=1-\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)^{2}=\frac{16-8-2 \sqrt{12}}{16}=\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^{2}
$$

Then, $\sin \alpha=\frac{\sqrt{6}-\sqrt{2}}{4}$, and

$$
\sin \alpha \cdot \cos \alpha=\frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}=\frac{6-2}{16}=\frac{1}{4}
$$

Note: If the student obtains $\sin \alpha=\frac{\sqrt{2-\sqrt{3}}}{2}$, to get the final solution (and the last point), the following procedure should be carried out:

$\sin \alpha \cdot \cos \alpha=\frac{\sqrt{2-\sqrt{3}}}{2} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\sqrt{(2-\sqrt{3})(\sqrt{6}+\sqrt{2})^{2}}}{8}=\frac{\sqrt{(2-\sqrt{3})(8+4 \sqrt{3})}}{8}=\frac{1}{4}$.