1/2sin(540+betta)*sin(betta+810)= sin 3alfa cos 2alfa+sin - вопрос №3226521125408103223 от 5473894 02.05.2020 07:50

Question

Алгебра: 1/2sin(540+betta)*sin(betta+810)= sin 3alfa cos 2alfa+sin 2alfa cos 3alfa- cos(2п-alfa)= sin(-alfa)+cos(п+alfa) / 1+2cos(п/2-alfa) cos(-alfa)= 4sin 10 cos 50 cos40= 1-cos2alfa / sin 2alfa

5473894 · Answer

1)1/2sin(540+b)*sin(b+810)=1/2sin(180+b)*sin(b+90)=1/2*(-sinb)*cosb=-1/sin2b
2)sin3acos2a+sin2acos3a-cos(2π-a)=sin(3a+2a)-cosa=sin5a-cosa
3)sin(-a)+cos(π+a)/1+2cos(π/2-a)cos(-a)=-sina-cosa/1+2sinacosa=
=-(sina+cosa)/(sina+cosa)²=-1/(sina+cosa)
4)4sin10*cos50*cos40=4sin10*cos50*sin50=2sin10*sin100=2sin10*sin(90+10)=
=2sin10*cos10=sin20
5)1-cos2a/sin2a=2sin²a/2sinacosa=sina/cosa=tga