Question
Prove that:
$\sin3\text{A}+\sin2\text{A}-\sin\text{A}=4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}$
$\sin3\text{A}+\sin2\text{A}-\sin\text{A}=4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}$
✓
Answer
We have,
$\text{LHS}=\sin3\text{A}+\sin2\text{A}-\sin\text{A}$
$=\ \sin3\text{A}-\sin\text{A}+\sin2\text{A}$
$=\ 2\sin\Big(\frac{3\text{A}-\text{A}}{2}\Big)\cos\Big(\frac{3\text{A}+\text{A}}{2}\Big)+\sin2\text{A}$
$=\ 2\sin\text{A}\cos2\text{A}+\sin2\text{A}$
$=\ 2\sin\text{A}\cos2\text{A}+2\sin\text{A}\cos\text{A}$
$=\ 2\sin\text{A}[\cos2\text{A}+\cos\text{A}]$
$=\ 2\sin\text{A}\Big[2\cos\Big(\frac{2\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{2\text{A}-\text{A}}{2}\Big)\Big]$
$=\ 4\sin\text{A}\cos\frac{3\text{A}}{2}\cos\frac{\text{A}}{2}$
$=\ 4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}$
$=\ \text{RHS}$
$\therefore\ \sin3\text{A}+\sin2\text{A}-\sin\text{A}=4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}.$
Hence proved.
$\text{LHS}=\sin3\text{A}+\sin2\text{A}-\sin\text{A}$
$=\ \sin3\text{A}-\sin\text{A}+\sin2\text{A}$
$=\ 2\sin\Big(\frac{3\text{A}-\text{A}}{2}\Big)\cos\Big(\frac{3\text{A}+\text{A}}{2}\Big)+\sin2\text{A}$
$=\ 2\sin\text{A}\cos2\text{A}+\sin2\text{A}$
$=\ 2\sin\text{A}\cos2\text{A}+2\sin\text{A}\cos\text{A}$
$=\ 2\sin\text{A}[\cos2\text{A}+\cos\text{A}]$
$=\ 2\sin\text{A}\Big[2\cos\Big(\frac{2\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{2\text{A}-\text{A}}{2}\Big)\Big]$
$=\ 4\sin\text{A}\cos\frac{3\text{A}}{2}\cos\frac{\text{A}}{2}$
$=\ 4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}$
$=\ \text{RHS}$
$\therefore\ \sin3\text{A}+\sin2\text{A}-\sin\text{A}=4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}.$
Hence proved.
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