[tex] \frac{sin(60 + \alpha ) - \frac{1}{2} sin \alpha }{ \cos(60 - \alpha ) - \frac{1}{2} \cos( \alpha ) } [/tex]. Created by nastyakrinitsina2005. algebra-ru

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nastyakrinitsina2005 @nastyakrinitsina2005

March 2022 1 66 Report

[tex] \frac{sin(60 + \alpha ) - \frac{1}{2} sin \alpha }{ \cos(60 - \alpha ) - \frac{1}{2} \cos( \alpha ) } [/tex]

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DimaPuchkov

Ответ:

$\frac{\sin{(60^o+\alpha)}-\frac{1}{2}\sin{\alpha}}{\cos{(60^o-\alpha)}-\frac{1}{2}\cos{\alpha}} = \frac{\sin{60^o}\cdot\cos{\alpha}+\cos{60^o}\cdot\sin{\alpha}-\frac{1}{2}\sin{\alpha}}{\cos{60^o}\cdot\cos{\alpha}+\sin{60^o}\cdot\sin{\alpha}-\frac{1}{2}\cos{\alpha}}=\frac{\frac{\sqrt{3}}{2}\cdot\cos{\alpha}+\frac{1}{2}\cdot\sin{\alpha}-\frac{1}{2}\sin{\alpha}}{\frac{1}{2}\cdot\cos{\alpha}+\frac{\sqrt{3}}{2}\cdot\sin{\alpha}-\frac{1}{2}\cos{\alpha}}=$

$=\frac{\frac{\sqrt{3}}{2}\cdot \cos{\alpha}}{\frac{\sqrt{3}}{2}\cdot\sin{\alpha}}=ctg\alpha$

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