Question

29.7. Найдите значение тригонометрического выражения: 1) cos11a + 3cos9a + 3 cos7a + cos 5a / cos 8 a если cosa = 1/3​

Answer 1

Упростим сначала числитель дроби :

$\displaystyle\bf\\1) \ \ Cos11\alpha +3Cos9\alpha +3Cos7\alpha +Cos5\alpha =\\\\=(Cos11\alpha +Cos5\alpha #)+3\cdot(Cos9\alpha +Cos7\alpha)=\\\\=2Cos\frac{11\alpha +5\alpha }{2} Cos\frac{11\alpha -5\alpha }{2} +3\cdot2Cos\frac{9\alpha +7\alpha }{2} Cos\frac{9\alpha -7\alpha }{2} =\\\\\\=2Cos8\alpha Cos3\alpha +6Cos8\alpha Cos\alpha =2Cos8\alpha \cdot(Cos3\alpha +3Cos\alpha )\\\\\\\ 2) \ \frac{Cos11\alpha +3Cos9\alpha +3Cos7\alpha +Cos5\alpha }{Cos8\alpha } =$

$\displaystyle\bf\\=\frac{2Cos8\alpha \cdot(Cos3\alpha +3Cos\alpha )}{Cos8\alpha } =2(Cos3\alpha +3Cos\alpha )\\\\\\Cos\alpha =\frac{1}{3} \\\\\\2(Cos3\alpha +3Cos\alpha )=2(Cos3\alpha +3\cdot \frac{1}{3} )=2(Cos3\alpha +1)=\\\\\\=2\cdot(4Cos^{3} \alpha -3Cos\alpha +1)=2\cdot\Big[4\cdot\Big(\frac{1}{3} \Big)^{3} -3\cdot\frac{1}{3} +1\Big]=\\\\\\=2\cdot\Big(4\cdot\frac{1}{27} -1+1\Big)=\frac{8}{27}$