Question

Упроситить
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

Answer 1

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