Question

Докажите тождество пожалуйста 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

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

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