Задача 69018 ...
Условие
2) (5cos x + 4)/(10sin x – 1) = ?
Решение
sin(π/12)*cos(π/4) - cos(π/12)*sin(π/4) = sin(π/12 - π/4) =
= sin(π/12 - 3π/12) = sin(-2π/12) = sin(-π/6) = -1/2
[b]Ответ: -1/2[/b]
2) tg(x/2) = 2
(5cos x + 4)/(10sin x - 1) = ?
Через формулы половинного аргумента:
[m]tg^2(x/2) = \frac{1-cos(x)}{1+cos(x)} = -\frac{cos(x)-1}{cos(x)+1} = -\frac{cos(x)+1-2}{cos(x)+1}=-1+\frac{2}{cos(x)+1}[/m]
[m]\frac{2}{cos(x)+1} = 1+tg^2(x/2) = 1+2^2=5[/m]
[m]1+cos(x) = \frac{2}{5}[/m]
[m]cos(x) = \frac{2}{5} -1 = -\frac{3}{5}[/m]
[m]sin(x) = \sqrt{1-cos^2(x)} = \sqrt{1-\frac{9}{25}} = \frac{16}{25}[/m]
[m]sin(x) = \frac{4}{5}[/m]
[m]\frac{5cos(x) + 4}{10sin(x) - 1} = \frac{5(-3/5) + 4}{10 \cdot (4/5) - 1} = \frac{-3 + 4}{8 - 1} = \frac{1}{7}[/m]
[b]Ответ: 1/7[/b]