Question

помогите пожолста))))

Answer 1

$\sin^2\alpha+\cos^2\alpha=1\\\sin^2\alpha=1-\cos^2\alpha\\\sin\alpha=\pm\sqrt{1-\cos^2\alpha}\\\\\sin\alpha=\pm\sqrt{1-\left(\frac{\sqrt{15}}4\right)^2}=\pm\sqrt{1-\frac{15}{16}}=\pm\sqrt{\frac1{16}}=\pm\frac14\\\\270^o<\alpha<360^o\Rightarrow\sin\alpha<0\\\\\boxed{\sin\alpha=-\frac14}$