Prove that: `sinA.sin(60^(@)+A).sin(60^(@)-A)=1/4sinA`

1.	Prove that: `sinA.sin(60^(@)+A).sin(60^(@)-A)=1/4sinA`
Answer» LHS `=sinA.sin(60^(@)+A).sin(60^(@)-A)` `=1/2sinA.[2sin(60^(@)+A)-(60^(@)-A)]` `1/2sinA[cos{(60^(@)+A)-(60^(@)-A)}]-cos{(60^(@)+A)+(60^(@)-A)}` `=1/2sinA[cos2A-cos120^(@)]` `=1/2[cos2AsinA-cos(90^(@)+30^(@)).sinA]` `=1/4[2cos2AsinA+2sin30^(@)sinA]` `=1/4[sin(2A+A)-sin(2A-A)=2.1/2.sinA]` `=1/4[sin3A-sinA+sinA]` `=14sin3A=R.H.S` Hence Proved.

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