\sqrt2\cos(\alpha-\beta),\;\cos\alpha\cos\beta=\frac12,\;a+b=\frac{5\pi}2c\mathrm{os}\alpha\cos\beta=\frac12(\;\cos(\alpha+b)+\cos(\alpha-b))\frac12=\frac12(\;\cos(\alpha+b)+\cos(\alpha-b))1=\;\cos(\alpha+b)+\cos(\alpha-b)1-\cos(\alpha+b)=\cos(\alpha-b)\cos(\alpha+b)=\cos(\frac{5\pi}2)=\cos(2\pi+\frac\pi2)=-\cos\frac\pi2=-01+0=\cos(\alpha-b)\cos(\alpha-b)=1\sqrt2\cos(\alpha-b)=\sqrt2Ответ: √2