Question

как решить, помогите пожалуйста ​

Answer 1

Ответ:

$\displaystyle tg\frac{x}{2}\approx 2,18$

Объяснение:

дано:

$\displaystyle cosx=-\frac{15}{23},~~~x\in\bigg(\frac{\pi }{2} ;\pi \bigg)\in(90^\circ;180^\circ)$

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$\displaystyle tg\frac{x}{2} =?$

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решение:

$\displaystyle sinx=\sqrt{1-cos^2x} =\sqrt{1-\bigg(-\frac{15}{23}\bigg)^2 }=\sqrt{1^{(529}-\frac{225}{529} } =\sqrt{\frac{529-225}{529} } =\sqrt{\frac{304}{529} } =\frac{4\sqrt{19} }{23} ;\\\displaystyle tg\frac{x}{2} =\frac{1-cosx}{sinx} =\frac{\displaystyle1-\bigg(-\frac{15}{23} \bigg)}{\displaystyle \frac{4\sqrt{19} }{23} } =\frac{\displaystyle \frac{38}{23} }{\displaystyle \frac{4\sqrt{19} }{23} } =\frac{38}{4\sqrt{19} } =\frac{\sqrt{19} *\sqrt{19} }{2\sqrt{19} } =\frac{\sqrt{19} }{2} \approx 2,18$