Докажите тригонометрическое тождество. Даю 100 баллов, срочно. [tex]\frac{\sqrt{2} cosa-2cos(\frac{\pi }{4} -a)}{2sin(\frac{\pi }{6} +a) -\sqrt{3}sina} =-\sqrt{2} tga[/tex]. Created by generatorenotov. algebra-ru

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NNNLLL54

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Ответ:

$\displaystyle \frac{\sqrt2\, cosa-2\, cos\Big(\dfrac{\pi}{4}-a\Big)}{2sin\Big(\dfrac{\pi}{6}+a\Big)-\sqrt3sina}=\frac{\sqrt2cosa-2\Big(cos\dfrac{\pi}{4}\cdot cosa+sin\dfrac{\pi}{4}\cdot sina\Big)}{2\Big(sin\dfrac{\pi}{6}\cdot cosa+sina\cdot cos\dfrac{\pi}{6}\Big)-\sqrt3sina}=\\\\\\=\frac{\sqrt2\, cosa-2\cdot \dfrac{\sqrt2}{2}\, cosa-2\cdot \dfrac{\sqrt2}{2}sina}{2\cdot \dfrac{1}{2}cosa+2\cdot \dfrac{\sqrt3}{2}\, sina-\sqrt3sina}=\frac{\sqrt2\, cosa-\sqrt2\, cosa-\sqrt2sina}{cosa+\sqrt3sina-\sqrt3sina}=$

$=\dfrac{-\sqrt2sina}{cosa}=-\sqrt2\cdot \dfrac{sina}{cosa}=-\sqrt2\, tga\\\\\\-\sqrt2\, tga=-\sqrt2\, tga$

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sangers1959

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Объяснение:

$\frac{\sqrt{2}*cos\alpha -2*cos(\frac{\pi }{4} -a) }{2*sin(\frac{\pi }{6}+\alpha)-\sqrt{3}*sin\alpha }.$

Упростим числитель:

$\sqrt{2}*cos\alpha -2*cos(\frac{\pi }{4}-\alpha )=\sqrt{2}*cos\alpha -2*(cos\frac{\pi }{2} *cos\alpha +sin\frac{\pi }{4}sin\alpha)=\\=\sqrt{2}*cos\alpha -2*(\frac{\sqrt{2} }{2}*cos\alpha +\frac{\sqrt{2} }{2}*sin\alpha)=\sqrt{2}*cos\alpha -\sqrt{2}*cos\alpha-\sqrt{2}*sin\alpha =\\=-\sqrt{2} *sin\alpha .$

Упростим знаменатель:

$2*sin(\frac{\pi }{6}+\alpha)-\sqrt{3}*sin\alpha = 2*(sin\frac{\pi }{6}*cos\alpha +cos\frac{\pi }{6} *sin\alpha)-\sqrt{3}*sin\alpha =\\=2*(\frac{cosx}{2} +\frac{\sqrt{3} }{2}*sin\alpha )-\sqrt{3}*sin\alpha =cosx+\sqrt{3}*sin\alpha-\sqrt{3}*sin\alpha =cosx.$

Таким образом:

$\frac{-\sqrt{2}*sinx }{cosx} =-\sqrt{2} tg\alpha .$

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