Упростите выражение. С решением пожалуйста . Created by razabovtimur76. algebra-ru

Упростите выражение. С решением пожалуйста ...

AlyaksandrK

$1)Cos^{3}\alpha Sin\alpha -Sin^{3}\alpha Cos\alpha=Sin\alpha Cos\alpha(Cos^{2}\alpha-Sin^{2}\alpha)=Sin\alpha Cos\alpha Cos2\alpha=\\\\=0,5*\underbrace{2Sin\alpha Cos\alpha}_{Sin2\alpha}*Cos2\alpha=0,5Sin2\alpha Cos2\alpha =\\\\=0,25*2Sin2\alpha Cos2\alpha=\boxed{0,25Sin4\alpha}\\\\\\2)\frac{Sin\alpha+Sin2\alpha}{1+Cos\alpha+Cos2\alpha}=\frac{Sin\alpha+2Sin\alpha Cos\alpha}{(1+Cos2\alpha)+Cos\alpha}=\frac{Sin\alpha(1+2Cos\alpha)}{2Cos^{2}\alpha+Cos\alpha}=$

$=\frac{Sin\alpha(1+2Cos\alpha)}{Cos\alpha(2Cos\alpha +1)}=\frac{Sin\alpha }{Cos\alpha}=\boxed{tg\alpha}$

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Universalka

Verified answer

Ответ: 1) $\frac{1}{4} sin4\alpha$ ; 2) $tg\alpha$

Объяснение:

$cos^{3} \alpha sin\alpha -sin^{3} \alpha cos\alpha =cos\alpha sin\alpha (cos^{2} \alpha -sin^{2} \alpha )=\frac{1}{2} 2cos\alpha sin\alpha cos2\alpha =\\=\frac{1}{2} sin2\alpha cos2\alpha =\frac{1}{4} 2sin2\alpha cos2\alpha=\frac{1}{4} sin4\alpha$

$\frac{sin\alpha +sin2\alpha }{1+cos\alpha +cos^{2}2\alpha } =\frac{sin\alpha +2sin\alpha cos\alpha }{cos^{2}\alpha +sin^{2}\alpha +cos\alpha +cos^{2}\alpha -sin^{2}\alpha } =\\\\=\frac{sin\alpha (1+2cos\alpha )}{2cos^{2}\alpha +cos\alpha } =\frac{sin\alpha (1+2cos\alpha )}{cos\alpha (2cos\alpha +1) }=\frac{sin\alpha }{cos\alpha } =tg\alpha$

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