7.2.1. (12 points) Calculate the value of the expression $\arccos \frac{\sqrt{6}+1}{2 \sqrt{3}}-\arccos \sqrt{\frac{2}{3}}$. Write the obtained expression in the form $\frac{a \pi}{b}$, where $a$ and $b$ are integers, and are coprime, and specify the value of $|a-b|$.

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# Answer: 7.

Solution. Since $\alpha=\arccos \frac{\sqrt{6}+1}{2 \sqrt{3}} \in\left(0 ; \frac{\pi}{2}\right)$ and $\beta=\arccos \sqrt{\frac{2}{3}} \in\left(0 ; \frac{\pi}{2}\right)$, then $A=\alpha-\beta \in\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$. Therefore, $\sin A=\sin \alpha \cos \beta-\cos \alpha \sin \beta$. After calculations, we get $\sin A=-\frac{1}{2}$, from which it follows that $A=-\frac{\pi}{6}$. Thus, $a=-1, b=6$, and $|a-b|=7$. Note that the answer can also be written as $a=1, b=-6$, but the absolute value of the difference remains the same.