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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\sin\text{A}+\sin2\text{A}+\sin4\text{A}+\sin5\text{A}=4\cos\frac{\text{A}}{2}\cos\frac{2\text{A}}{2}\cos4\text{A}$
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Answer

We have,
$\text{LHS}=\sin\text{A}+\sin2\text{A}+\sin4\text{A}+\sin5\text{A}$
$=\ (\sin2\text{A}+\sin\text{A})+(\sin5\text{A}+\sin4\text{A})$
$=\ \Big[2\sin\Big(\frac{2\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{2\text{A}-\text{A}}{2}\Big)\Big]+\Big[2\sin\Big(\frac{5\text{A}+4\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-4\text{A}}{2}\Big)\Big]$
$=\ 2\sin\frac{3\text{A}}{2}\cos\frac{\text{A}}{2}+2\sin\frac{9\text{A}}{2}\cos\frac{\text{A}}{2}$
$=\ 2\cos\frac{\text{A}}{2}\Big[\sin\frac{3\text{A}}{2}+\sin\frac{9\text{A}}{2}\Big]$
$=\ 2\cos\frac{\text{A}}{2}\Big[\sin\frac{9\text{A}}{2}+\sin\frac{3\text{A}}{2}\Big]$
$=\ 2\cos\frac{\text{A}}{2}\Big[2\sin\Big\{\frac{1}{2}\Big(\frac{9\text{A}}{2}+\frac{3\text{A}}{2}\Big)\Big\}\cos\Big\{\frac{1}{2}\Big(\frac{9\text{A}}{2}-\frac{3\text{A}}{2}\Big)\Big\}\Big]$
$=\ 4\cos\frac{\text{A}}{2}\Big[\sin\frac{12\text{A}}{4}\cos\frac{6\text{A}}{4}\Big]$
$=\ 4\cos\frac{\text{A}}{2}\sin3\text{A}\cos\frac{3\text{A}}{2}$
$=\ 4\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}\sin3\text{A}$
$=\ \text{RHS}$
$\therefore\ \sin\text{A}+\sin2\text{A}+\sin4\text{A}+\sin5\text{A}=4\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}\sin3\text{A}.$
Hence proved.

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