Question

Prove that SinAsin (60°-A)sin (60°+A)=1/4sin3A

Answer 1

LHS=sinA(sin60-A)sin(60+A)
       =sinA[sin^2(60)-sin^2(A)]
       =sinA[3/4-sin^2(A)]
       =3sinA/4-sin^3(A)= 1/4[3sinA-4sin^3(A)]
       =1/4[sin3(A)]=RHS

oh and btw sin^2 means sine squared and sin^3 means sine cubed 
Your welcome homie

Answer 2

Answer:

Here is your answer buddy...