Question
Prove that:$\sin^2\text{B}=\sin^2\text{A}+\sin^2\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\sin\text{(A}-\text{B)} $
✓
Answer
$\text{R.H.S}=\sin^2\text{A}+\sin^2\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\sin\text{(A}-\text{B)} $
$=\sin^2\text{A}+\sin\text{(A}-\text{B)}\Big[\sin\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\Big] $
$=\sin^2\text{A}+\sin\text{(A}-\text{B)}\Big[-\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}\ \Big]$
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}\Big[\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\ \Big]$
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}(\sin(\text{A}+\text{B))} $
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}\sin(\text{A}+\text{B)} $
$=\sin^2\text{A}-\sin(\sin^2\text{A}-\sin^2\text{B})$
$=\sin^2\text{A}-\sin\sin^2\text{A}+\sin^2\text{B}$$\Big[\because\sin(\text{A}-\text{B)}\sin\text{(A+}\text{B)}=\sin^2\text{A}-\sin^2\text{B}\Big]$
$=\sin^2\text{B}$
$=\text{L.H.S}$
$\therefore \text{L.H.S}=\text{R.H.S}$
Hence proved.
$=\sin^2\text{A}+\sin\text{(A}-\text{B)}\Big[\sin\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\Big] $
$=\sin^2\text{A}+\sin\text{(A}-\text{B)}\Big[-\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}\ \Big]$
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}\Big[\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\ \Big]$
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}(\sin(\text{A}+\text{B))} $
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}\sin(\text{A}+\text{B)} $
$=\sin^2\text{A}-\sin(\sin^2\text{A}-\sin^2\text{B})$
$=\sin^2\text{A}-\sin\sin^2\text{A}+\sin^2\text{B}$$\Big[\because\sin(\text{A}-\text{B)}\sin\text{(A+}\text{B)}=\sin^2\text{A}-\sin^2\text{B}\Big]$
$=\sin^2\text{B}$
$=\text{L.H.S}$
$\therefore \text{L.H.S}=\text{R.H.S}$
Hence proved.
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