Question

Помогите пожалуйста!

Answer 1

1)sin105sin15=1/2[cos(105-15)-cos(105+15)]=1/2[cos90-cos(180-60)]=
=1/2(0+cos60)=1/2*1/2=1/4
2)sin(π/3-a)+sin(2π/3-a)=2sin(π/2-a)cos(-π/6)=2cosa*√3/2=√3cosa
3)2cos(a+b)cos(a-b)-1+2sin²b=2*1/2(cos2a+cos2b)-cos2b=cos2a+cos2b-cos2b=
=cos2a
4)(2sin3acosa-sin2a)/(cos2a-cos6a)=(2sin3acosa-2sinacosa)/2sin2asin4a=
=2cosa(sin3a-sina)/2sin2asin4a=cosa*2sinacos2a/4sinacosasin2acos2a=
=1/2sin2a=sina/4sinacosa