Задача 77251 ...

65e0bc85d453197203cf91a9

5/2/2024 4:07:58 PM

Дано: cosx = -20/29, если -π/2 < x < π

Найти: sinx -?; sin2x-?

ctg (3π/2 + x)-?

626fab9a354ab65fb2f43abb

5/2/2024 4:34:02 PM

★

cos x = -20/29, π/2 < x < π, значит, cos x < 0, sin x > 0
sin^2 x = 1 - cos^2 x = 1 - (-20/29)^2 = 1 - 400/841 = 441/841

sin x = +sqrt(441/841) = 21/29

[m]\sin 2x = 2\sin x \cdot \cos x = 2 \cdot \frac{21}{29}(-\frac{20}{29}) = - \frac{2 \cdot 21 \cdot 20}{29 \cdot 29} = -\frac{840}{841}[/m]

[m]ctg(\frac{3\pi}{2} + x) = -tg(x) = -\frac{\sin x}{\cos x} = -\frac{21}{29} : (-\frac{20}{29}) = \frac{21}{29} \cdot \frac{29}{20} = \frac{21}{20}[/m]