Question

2. Если tga = -√5 и π/2<a<π, найдите sin a​

Answer 1

$\displaystyle\bf\\\frac{\pi }{2} < \alpha < \pi \ \ \ \Rightarrow \ \ \ Sin\alpha > 0\\\\\\tg\alpha =-\sqrt{5} \ \ \ \ \Rightarrow \ \ \ Ctg\alpha =\frac{1}{tg\alpha } =-\frac{1}{\sqrt{5} } \\\\\\1+Ctg^{2} \alpha =\frac{1}{Sin^{2} \alpha } \\\\\\Sin^{2} \alpha =\frac{1}{1+Ctg^{2}\alpha } =\frac{1}{1+\Big(-\dfrac{1}{\sqrt{5} } \Big)^{2} } =\frac{1}{1+\dfrac{1}{5} } =\\\\\\=1:\frac{6}{5} =1\cdot \frac{5}{6} =\frac{5}{6} \\\\\\\boxed{Sin\alpha =\sqrt{\frac{5}{6} } }$