Question

помогите плез срочно оч надо ребят

Answer 1

$1)\; \; cosx=\frac{\sqrt{15}}{8}\\\\270^\circ <x<360^\circ \; \; \; \Rightarrow \; \; \; sinx<0\\\\\boxed {\; sin^2x+cos^2x=1\; }\; \; \; \Rightarrow \; \; \; \boxed {sin^2x=1-cos^2x\; }\\\\sin^2x=\pm \sqrt{1-cos^2x}\\\\sinx<0\; \; \to \; \; \; sinx=-\sqrt{1-cos^2x}=-\sqrt{1-\frac{15}{64}}=-\sqrt{\frac{49}{64}}=-\frac{7}{8}$

$2)\; \; sinx=\frac{\sqrt{39}}{8}\\\\90^\circ <x<180^\circ \; \; \; \Rightarrow \; \; \; cosx<0\\\\\boxed {\; cos^2x=1-sin^2x}\\\\cosx<0\; \; \to \; \; \; cosx=-\sqrt{1-sin^2x}=-\sqrt{1-\frac{39}{64}}=-\sqrt{\frac{25}{64}}=-\frac{5}{8}$