Question

Докажите тождество 50ballov 1) 4sin(п/6-y)*cos(п/6+y)=v3-2sin2y v - это корень 2) -4sin(п/6+y)*sin(п/6-y)=3-4cos^2 y

Answer 1

$1)4sin(\frac{\pi}{6}-y)*cos(\frac{\pi}{6}+y)=\sqrt{3}-2sin(2y)$
$4*(sin(\frac{\pi}{6}-y)*cos(\frac{\pi}{6}+y))=\sqrt{3}-2sin(2y)$
$4*(\frac{1}{2}(sin((\frac{\pi}{6}+y)+(\frac{\pi}{6}-y))-sin((\frac{\pi}{6}+y)-(\frac{\pi}{6}-y))))=$
$=\sqrt{3}-2sin(2y)$
$4*(\frac{1}{2}(sin(\frac{\pi}{3})-sin2y))=\sqrt{3}-2sin(2y)$
$2(sin\frac{\pi}{3}-sin2y)=\sqrt{3}-2sin(2y)$
$2(\frac{\sqrt{3}}{2}-sin2y)=\sqrt{3}-2sin(2y)$
$\sqrt{3}-2sin2y=\sqrt{3}-2sin2y$

$2)-4sin(\frac{\pi}{6}+y)*sin(\frac{\pi}{6}-y)=3-4cos^2y$
$-4*(\frac{1}{2}(cos((\frac{\pi}{6}-y)-(\frac{\pi}{6}+y))-cos((\frac{\pi}{6}-y)+(\frac{\pi}{6}+y))))=$
$=3-4cos^2y$
$-4*(\frac{1}{2}(cos(-2y)-cos(\frac{\pi}{3})))=3-4cos^2y$
$-2(cos(-2y)-\frac{1}{2})=3-4cos^2y$
$-2(cos(2y)-\frac{1}{2})=3-4cos^2y$
$1-2cos2y=3-4cos^2y$
$1-2cos2y=3-4*\frac{1+cos2y}{2}$
$1-2cos2y=3-2-2cos2y$
$1-2cos2y=1-cos2y$