Question

证明sinA+sin(A+2π/3)+sin(A-2π/3)=0

Answer 1

解：
sinA+sin(A+2π/3)+sin(A-2π/3)
=sinA+sinAcos2π/3+cosAsin2π/3+sinAcos2π/3-cosAsin2π/3
=sinA+2sinAcos2π/3
=sinA+2sinA*(-1/2)
=sinA-sinA
=0.

Answer 2

证明sinA sin(A 2π/3) sin(A-2π/3)=0
nA sin（A 2/3π） cos（A 5/6π）
=sina sinacos2/3pai sin2/3paicosa cosacos5/6pai-sinasin5/6pai
=sina-1/2sina 根号3/2cosa-根号3/2cosa-1/2sina
=0

Answer 3

(b-3)/(sin45-sin15)=4/sin45 因：sin45-sin15= 2cos30sin15 √2/2-sin15=√3sin15 sin15=(√6-√2)/4 所以：b=a/sinA*sinB =12√2/[

Answer 4

d