B1. Accurately, without using a pocket calculator, calculate the value of the expression

$$
\frac{\cos 20^{\circ}-\sin 650^{\circ}}{\sin \frac{\pi}{12} \cdot \cos \left(-520^{\circ}\right)}+2^{2} \cdot \cos \frac{\pi}{4}
$$

Rationalize the denominator.
B1. How do we express the trigonometric functions of any angles in terms of trigonometric functions of acute angles: $\sin 650^{\circ}=$ $\sin \left(650^{\circ}-2 \cdot 360^{\circ}\right)=\sin \left(-70^{\circ}\right)=-\sin 70^{\circ}=-\cos 20^{\circ}, \cos \left(-520^{\circ}\right)=\cos 520^{\circ}=\cos \left(520^{\circ}-\right.$ $\left.360^{\circ}\right)=\cos 160^{\circ}=-\cos 20^{\circ}$. For $\sin \frac{\pi}{2}$ we use the addition formula: $\sin \frac{\pi}{12}=\sin 15^{\circ}=\sin \left(45^{\circ}-\right.$ $\left.30^{\circ}\right)=\frac{\sqrt{6}-\sqrt{2}}{4}$. We calculate the second term $4 \cdot \frac{\sqrt{2}}{2}=2 \sqrt{2}$. We substitute the data into the expression and

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-17.jpg?height=83&width=1674&top_left_y=1763&top_left_x=228)
analyze the denominator $-\frac{2 \cdot 4}{\sqrt{6}-\sqrt{2}} \cdot \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}+2 \sqrt{2}=-\frac{8(\sqrt{6}+\sqrt{2})}{4}+2 \sqrt{2}$, simplify and get $-2 \sqrt{6}-2 \sqrt{2}+2 \sqrt{2}=-2 \sqrt{6}$.

The expression $\sin 650^{\circ}=\sin \left(650^{\circ}-2 \cdot 360^{\circ}\right)=\sin \left(-70^{\circ}\right)=-\sin 70^{\circ}=-\cos 20^{\circ} \ldots \ldots \ldots \ldots .1$ point The expression $\cos \left(-520^{\circ}\right)=\cos 520^{\circ}=\cos \left(520^{\circ}-360^{\circ}\right)=\cos 160^{\circ}=-\cos 20^{\circ} \ldots \ldots \ldots \ldots \ldots . \ldots$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-17.jpg?height=70&width=1674&top_left_y=2098&top_left_x=228)

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-17.jpg?height=68&width=1674&top_left_y=2156&top_left_x=228)

Substituted data $\frac{\cos 20^{\circ}+\cos 20^{\circ}}{-\cos 20^{\circ} \cdot \frac{\sqrt{6}-\sqrt{2}}{4}+2 \sqrt{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 1$ point

Rationalization of the denominator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $1^{*}$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-17.jpg?height=57&width=1670&top_left_y=2336&top_left_x=233)