Prove that: 3A+ 2A- =4 A 2 3A 2 - Vidyadip

Question

Prove that: $\sin3\text{A}+\sin2\text{A}-\sin\text{A}=4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}$

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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.

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