Question

Найдите sin(a/2), если sin(a) = 24/25, 0<a<π
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Answer 1

sin(a) = 2·sin(a/2)·cos(a/2) = 24/25

sin(a/2)·cos(a/2) = 12/25,

если 0<a<π, то 0<a/2<π/2 и cos(a/2) > 0, тогда

$\cos(\frac{a}{2}) = \sqrt{1 - \sin^2(a/2)}$

$\sin(\frac{a}{2})\cdot\sqrt{1 - \sin^2(a/2)} = \frac{12}{25}$

sin(a/2) = x,

x > 0, т.к. 0<a/2<π/2

$x\cdot\sqrt{1 - x^2} = \frac{12}{25}$

возведем в квадрат

$x^2\cdot(1 - x^2) = \frac{12^2}{25^2}$

x² = t,

$t\cdot(1 - t) = \frac{12^2}{25^2}$

$t - t^2 = \frac{12^2}{25^2}$

$t^2 - t + \frac{12^2}{25^2} = 0$

$D = 1^2 - 4\cdot\frac{12^2}{25^2} = 1 - \frac{4\cdot 144}{625} =$

$= \frac{625 - 576}{625} = \frac{49}{625} = (\frac{7}{25})^2$

$t = \frac{1\pm\frac{7}{25}}{2}$

$t_1 = \frac{25 - 7}{50} = \frac{18}{50} = \frac{9}{25}$

$t_2 = \frac{25+7}{50} = \frac{32}{50} = \frac{16}{25}$

1. $x^2 = \frac{9}{25}$

$x > 0$

$x = \sqrt{\frac{9}{25}} = \frac{3}{5}$

2. $x^2 = \frac{16}{25}$

$x > 0$

$x = \sqrt{\frac{16}{25}} = \frac{4}{5}$