Question

Найдите значение выражения: 5cos(a - b), если cosa*cosb = 1/2; a + b = п/3.
4)4sin(a-b),если sina cosb=1/4;a+b=-π/6​

Answer 1

$\displaystyle\bf\\1)\\\\\alpha +\beta =\frac{\pi }{3} \\\\Cos\alpha Cos\beta =\frac{1}{2} \\\\\\Cos\alpha Cos\beta =\frac{1}{2} \bigg[Cos(\alpha -\beta )+Cos(\alpha +\beta )\bigg]=\frac{1}{2} \bigg[Cos(\alpha -\beta )+Cos\frac{\pi }{3} \bigg]=\\\\\\=\frac{1}{2}\bigg[Cos(\alpha -\beta )+\frac{1}{2} \Bigg] \\\\\\\frac{1}{2} =\frac{1}{2} \bigg[Cos(\alpha -\beta )+\frac{1}{2} \bigg]\\\\\\Cos(\alpha -\beta )+\frac{1}{2} =1\\\\\\ \ Cos(\alpha -\beta )=-\frac{1}{2} \\\\\\\boxed{5Cos(\alpha -\beta )=-2,5}$

$\displaystyle\bf\\2)\\\\\alpha +\beta =-\frac{\pi }{6} \\\\\\Sin\alpha Cos\beta =\frac{1}{4} \\\\\\Sin\alpha Cos\beta =\frac{1}{2} \bigg[Sin(\alpha -\beta )+Sin(\alpha +\beta )\bigg]=\frac{1}{2} \bigg[Sin(\alpha -\beta )+Sin(-\frac{\pi }{6}) \bigg]=\\\\\\=\frac{1}{2}\bigg[Sin(\alpha -\beta )-\frac{1}{2} \Bigg] \\\\\\\frac{1}{4} =\frac{1}{2} \bigg[Sin(\alpha -\beta )-\frac{1}{2} \bigg]\\\\\\Sin(\alpha -\beta )-\frac{1}{2} =\frac{1}{2} \\\\\\Sin(\alpha -\beta )=1 \\\\\ \boxed{4Sin(\alpha -\beta )=4}$