докажите тождество..... Created by kefgekiz. algebra-ru

докажите тождество.......

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$2sin^2(3\pi -2a)\cdot cos^2(5\pi +2a)=2sin^2(2\pi +\pi -2a)\cdot cos^2(4\pi +\pi +2a)=\\\\=2sin^2(\pi -2a)\cdot cos^2(\pi +2a)=2sin^22a\cdot cos^22a=\\\\=2\cdot (sin2a\cdot cos2a)^2=2\cdot (\dfrac{1}{2}\cdot sin4a)^2=\dfrac{1}{2}\cdot sin^24a=\\\\=\dfrac{1}{2}\cdot \dfrac{1-cos8a}{2}=\dfrac{1}{4}-\dfrac{1}{4}\cdot cos8a=\dfrac{1}{4}-\dfrac{1}{4}\cdot sin(\dfrac{\pi}{2}-8a)=\\\\=\dfrac{1}{4}-\dfrac{1}{4}\cdot sin(2\pi +\dfrac{\pi}{2}-8a)=\dfrac{1}{4}-\dfrac{1}{4}\cdot sin(\dfrac{5\pi}{2}-8a)$

$\star \ \ sin^2x=\dfrac{1-cos2x}{2}\ \ ,\ \ 2sinx\cdot cosx=sin2x\ \ ,\ \ cos(2\pi \pm x)=cosx\ ,\\\\cosx=sin(\dfrac{\pi}{2}-x)\ \ \star$

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OneGyrus

Добрый день!

$2sin^2(3\pi -2\alpha )*cos^2(5\pi +2a) = 2*sin^2(2a)*(-cos(2a) )^2 = \\=\frac{1}{2} *(2*sin(2a)*cos(2a) )^2 = \frac{1}{2} *sin^2(4a) =\\ = \frac{1-cos(8a)}{4} = \frac{1-sin(\frac{\pi }{2} -8a)}{4} =\frac{1}{4} -\frac{1}{4}*(sin(\frac{\pi }{2}+2\pi -8a) ) =\frac{1}{4} -\frac{1}{4}*(sin(\frac{5\pi }{2} -8a) )$ Тождество доказано!

Формулы применяемые в решении:

$sin(2a)=2*sin(a)*cos(a) \\sin^2(a) = \frac{1-cos(2a)}{2}$

Если решение понравилось, ставь лучший ответ!

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