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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Prove that:
$\sin3\text{x}+\sin2\text{x}-\sin\text{x}=4\sin\text{x}\cos\Big(\frac{\text{x}}{2}\Big)\cos\Big(\frac{3\text{x}}{2}\Big)$
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Answer

$\text{L.H.S.}=\sin3\text{x}+\sin2\text{x}-\sin\text{x}$$=\sin3\text{x}+(\sin2\text{x}-\sin\text{x})$
$=\sin3\text{x}+\Big[2\cos\Big(\frac{2\text{x}+\text{x}}{2}\Big)\sin\Big(\frac{2\text{x}-\text{x}}{2}\Big)\Big]$$\Big[\sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\sin3\text{x}+\Big[2\cos\Big(\frac{3\text{x}}{2}\Big)+\sin\Big(\frac{\text{x}}{2}\Big)\Big]$
$=\sin3\text{x}+2\cos\frac{3\text{x}}{2}\sin\frac{\text{x}}{2}$
$=2\sin\frac{3\text{x}}{2}.\cos\frac{3\text{x}}{2}+2\cos\frac{3\text{x}}{2}\sin\frac{\text{x}}{2}$ $[\sin2\text{A}=2\sin\text{A}.\cos\text{B}]$
$=2\sin\cos\Big(\frac{3\text{x}}{2}\Big)\Big[\sin\Big(\frac{3\text{x}}{2}\Big)+\sin\Big(\frac{\text{x}}{2}\Big)\Big]$
$=2\cos\Big(\frac{3\text{x}}{2}\Big)\begin{bmatrix}2\sin\begin{Bmatrix}\frac{\Big(\frac{3\text{x}}{2}\Big)+\Big(\frac{\text{x}}{2}\Big)}{2}\end{Bmatrix}\end{bmatrix}\cos\begin{Bmatrix}\frac{\Big(\frac{3\text{x}}{2}\Big)-\Big(\frac{\text{x}}{2}\Big)}{2}\end{Bmatrix}$ $\Big[\sin\text{A}+\sin\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=2\cos\Big(\frac{3\text{x}}{2}\Big).2\sin\text{x}\cos\Big(\frac{\text{x}}{2}\Big)$
$=4\sin\text{x}\cos\Big(\frac{\text{x}}{2}\Big)\cos\Big(\frac{3\text{x}}{2}\Big)=\text{R.H.S.}$

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