Question

Тригонометрия, вопрос: $\sqrt{6} * sin a$, если $cos a = 0,2$ и a относится (pi/2;2pi)

Answer 1

$cosa=0,2\\\\a\in \Big(\dfrac{\pi}{2}\, ;\, 2\pi \, \Big)\ \ \to \\\\\\esli\ \ a\in \Big(\dfrac{\pi}{2}\, ;\, \dfrac{3\pi }{2}\, \Big)\ ,\ to \ \ cosa<0\ \ i\ \ \ cosa\ne +0,2\\\\esli\ \ a\in \Big(\dfrac{3\pi}{2}\, ;\, 2\pi \, \Big)\ ,\ to\ \ cosa>0\ \ i\ \ cosa=+0,2\ \ ,\ \ sina<0\\\\sina=-\sqrt{1-cos^2a}=-\sqrt{1-0,04}=-\sqrt{0,96}=-\sqrt{\dfrac{96}{100}}=-\dfrac{4\sqrt6}{10}=-\dfrac{2\sqrt6}{5}\\\\\\\sqrt6\cdot sina=-\dfrac{2\sqrt6\cdot \sqrt6}{5}=-\dfrac{12}{5}=-2,4$