Помогите, пожалуйста, решить.

1) $39 \sqrt{2}*tg( \frac{\pi}{4} )*sin( \frac{\pi}{4} )= 39 \sqrt{2}* \frac{sin( \frac{\pi}{4})}{cos(\frac{\pi}{4})} * \frac{ \sqrt{2} }{2} =$

$=39 \sqrt{2}* \frac{ \frac{ \sqrt{2} }{2} }{ \frac{ \sqrt{2} }{2} } * \frac{ \sqrt{2} }{2} =39 \sqrt{2}*1*\frac{ \sqrt{2} }{2} =\frac{39* (\sqrt{2})^2 }{2} = \frac{39*2}{2} =39$

2) $\frac{48}{sin(- \frac{19\pi}{3} )cos( \frac{25\pi}{6} )} = \frac{48}{-sin(\frac{3*3*2\pi+\pi}{3} )cos( \frac{2*6*2\pi+\pi}{6} )} =$

$=- \frac{48}{sin(3*2\pi+\frac{\pi}{3} )cos( 2*2\pi+\frac{\pi}{6} )} =- \frac{48}{sin(\frac{\pi}{3} )cos(\frac{\pi}{6} )} =- \frac{48}{ \frac{ \sqrt{3} }{2} *\frac{ \sqrt{3} }{2} }=$

$=- \frac{48}{ \frac{3}{4} }=- \frac{48*4}{3} =- \frac{3*16*4}{3}=-16*4 -64$

3) $42 \sqrt{3}*sin(120^0)=42 \sqrt{3} *sin( \frac{\pi}{2}+ \frac{\pi}{6} )=42 \sqrt{3}*cos( \frac{\pi}{6} )=$

$=42 \sqrt{3}* \frac{ \sqrt{3} }{2}=21*3=63$