sina+cosa的和差化积

Question

积化和差和和差化积
在三角形ABC中,cosAcosB+sinAsinB+cosBsinA+cosAsinB=2,则三角形ABC是_____三角形
化简1/2[sin2θ+cos(π/3-2θ)]-sin（π/12）cos(π/12+2θ)

Answer 1

cosAcosB+sinAsinB+cosBsinA+cosAsinB=2
即cos(A-B)+sin(A+B)=2
因为cos(A-B)<=1
sin(A+B)<=1
所以cos(A-B)+sin(A+B)<=2
所以cos(A-B)=1
sin(A+B)=1
所以A-B=0
A+B=90
A=B=45
所以是等腰直角三角形
1/2[sin2θ+cos(π/3-2θ)]-sin（π/12）cos(π/12+2θ)
=1/2(sin2θ+sin(2θ+π/6))-1/2(sin(π/12+π/12+2θ)+sin(π/12-π/12-2θ)
=1/2(sin2θ+sin(2θ+π/6))-1/2(sinπ/6+2θ)+sin2θ/2
=sin2θ