Prove that: `sin(A+2B)sinA-sinBsin(2A+B)sinB=sin(A+B)sin(A-B)`

1.	Prove that: `sin(A+2B)sinA-sinBsin(2A+B)sinB=sin(A+B)sin(A-B)`
Answer» LHS. `=sin(A+2B)sinA-sin(2A+B)sinB` `=1/2[{cos(A+2B-A)-cos(A+2B+A)}-{cos(2A+B-B)-cos(2A+B+B)}]` `=1/2[cos2B-cos(2A+2B)-cos2A+cos(2A+2B)]` `=1/2(cos2B-cos2A)` and RHS. `=sin(A+B)sin(A-B)` `=1/2.2sin(A+B)sin(A-B)` `=1/2[cos{(A+B)-(A-B)}-cos{(A+B)+(A-B)}]` `1/2(cos2B-cos2A)` `therefore` L.H.S=R.H.S Hence Proved.

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