1. It is known that $\sin (\alpha+\beta)=0.2$ and $\cos (\alpha-\beta)=0.3$. Calculate $\sin \left(\alpha+45^{\circ}\right) \cdot \sin (\beta+$ $\left.45^{\circ}\right)$.
Answer: 0.25.

Solution. Using the formula for the sine of the sum of angles, we have

$$
\sin \left(\varphi+45^{\circ}\right)=\frac{\sqrt{2}}{2}(\sin \varphi+\cos \varphi)
$$

and thus, $P=\sin \left(\alpha+45^{\circ}\right) \cdot \sin \left(\beta+45^{\circ}\right)=\frac{1}{2}(\sin \alpha+\cos \alpha)(\sin \beta+\cos \beta)$. Expanding the brackets and using the formula for the sine and cosine of the sum of angles again, we get

$$
P=\frac{1}{2}(\sin (\alpha+\beta)+\cos (\alpha-\beta))=0.25
$$

Criteria. The equality $P=\frac{1}{2}(\sin \alpha+\cos \alpha)(\sin \beta+\cos \beta)-1$ point is proven.