Question

Ученикам 10 класса ВОПРОС 17Н

Answer 1

$x\in (\frac{3\pi}{2};2\pi )\; \; \; \to \; \; \; \frac{x}{2}\in (\frac{3\pi}{4};\pi )\; \; \to \; \; cos\frac{x}{2}<0\; ,\; \; sin\frac{x}{2}>0\\\\cosx=\frac{5}{13}\\\\cos\frac{x}{2}=-\sqrt{\frac{1+cosx}{2}}=-\sqrt{\frac{1+\frac{5}{13}}{2}}=-\sqrt{\frac{18}{13\cdot 2}}=-\frac{3}{\sqrt{13}}\\\\sin\frac{x}{2}=+\sqrt{\frac{1-cosx}{2}}=\sqrt{\frac{1-\frac{5}{13}}{2}}=\sqrt{\frac{8}{13\cdot 2}}=\frac{2}{\sqrt{13}}\\\\sin\frac{x}{2} +cos\frac{x}{2}+1,4=\frac{2}{\sqrt{13}}-\frac{3}{\sqrt{13}}+1,4=1,4-\frac{1}{\sqrt{13}}=\frac{7}{5}-\frac{\sqrt{13}}{13}=\frac{91-5\sqrt{13}}{65}$