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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
Prove that:
$\sin\frac{\text{x}}{2}\sin\frac{7\text{x}}{2}+\sin\frac{3\text{x}}{2}\sin\frac{11\text{x}}{2}=\sin2\text{x}\sin5\text{x}.$
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Answer

We have,
$\text{LHS}=\sin\frac{\text{x}}{2}\sin\frac{7\text{x}}{2}+\sin\frac{3\text{x}}{2}\sin\frac{11\text{x}}{2}$
$=\ \frac{1}{2}\Big[2\sin\frac{7\text{x}}{2}\sin\frac{\text{x}}{2}+2\sin\frac{110}{2}\sin\frac{3\text{x}}{2}\Big]$
$=\ \frac{1}{2}\Big[\cos\Big(\frac{7\text{x}}{2}-\frac{\text{x}}{2}\Big)-\cos\Big(\frac{7\text{x}}{2}+\frac{\text{x}}{2}\Big)+\cos\Big(\frac{110}{2}-\frac{3\text{x}}{2}\Big)-\cos\Big(\frac{110}{2}+\frac{3\text{x}}{2}\Big)\Big]$
$=\ \frac{1}{2}\Big[\cos\frac{60}{2}-\cos\frac{80}{2}+\cos\frac{80}{2}-\cos\frac{140}{2}\Big]$
$=\ \frac{1}{2}[\cos3\text{x}-\cos4\text{x}+\cos4\text{x}-\cos7\text{x}]$
$=\ \frac{1}{2}[\cos3\text{x}-\cos7\text{x}]$
$=\ \frac{-1}{2}[\cos7\text{x}-\cos3\text{x}]$
$=\ \frac{-1}{2}\Big[-2\sin\Big(\frac{7\text{x}+3\text{x}}{2}\Big)\sin\Big(\frac{7\text{x}-3\text{x}}{2}\Big)\Big]$
$=\ \sin\frac{10\text{x}}{2}\sin\frac{4\text{x}}{2}$
$=\ \sin5\text{x}\sin2\text{x}$
$=\ \sin2\text{x}\sin5\text{x}$
$=\ \text{RHS}$
$\therefore\ \sin\frac{\text{x}}{2}\sin\frac{7\text{x}}{2}+\sin\frac{3\text{x}}{2}\sin\frac{110}{2}=\sin2\text{x}\sin5\text{x}.$
Hence proved.

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