Question

100 баллов даю ребята, помогите пожалуйста
Найти sin (a-b), если sin a =3/5 и П/2<a<П, sin b= - 4/5 и П<b<3П/2​

Answer 1

$\displaystyle\bf\\\frac{\pi }{2} < \alpha < \pi \ \ \Rightarrow \ \ Cos\alpha < 0\\\\\\\pi < \beta < \frac{3\pi }{2} \ \ \Rightarrow \ \ Cos\beta < 0\\\\\\Sin\alpha =\frac{3}{5} \\\\\\Cos\alpha =-\sqrt{1-Sin^{2} \alpha } =-\sqrt{1-\Big(\frac{3}{5}\Big)^{2} } =-\sqrt{1-\frac{9}{25} } =-\sqrt{\frac{16}{25} } =-\frac{4}{5} \\\\\\Sin\beta =-\frac{4}{5}$

$\displaystyle\bf\\Cos\beta =-\sqrt{1-Sin^{2} \beta } =-\sqrt{1-\Big(-\frac{4}{5}\Big)^{2} } =-\sqrt{1-\frac{16}{25} } =-\sqrt{\frac{9}{25} } =-\frac{3}{5} \\\\\\Sin(\alpha -\beta )=Sin\alpha Cos\beta -Cos\alpha Sin\beta=\frac{3}{5} \cdot\Big(-\frac{3}{5} \Big) -\Big(-\frac{4}{5} \Big)\cdot\Big(-\frac{4}{5}\Big)=\\\\\\=-\frac{9}{25} -\frac{16}{25} =-\frac{25}{25}=-1\\\\\\Otvet: \ Sin(\alpha -\beta )=-1$