Question

Помогите пожалуйста ! Срочно !

Answer 1

$\displaystyle\bf\\\frac{3\pi }{2} <\alpha <2\pi \ \ \Rightarrow \ \ Sin\alpha <0\\\\\\Sin\alpha =-\sqrt{1-Cos^{2}\alpha } =-\sqrt{1-\Big(-\frac{\sqrt{15} }{4} \Big)^{2} } =-\sqrt{1-\frac{15}{16} } =\\\\\\=-\sqrt{\frac{1}{16} } =-\frac{1}{4}\\\\\\7-8Sin\alpha=7-8\cdot \Big(-\frac{1}{4}\Big)=7+2=9\\\\\\Otvet:9$