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Transformation Formulae — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae5 Marks
Question
Prove that:
$\cos\text{x}\cos\frac{\text{x}}{2}-\cos3\text{x}\cos\frac{9\text{x}}{2}=\sin7\text{x}\sin8\text{x}.$
✓

Answer

Consider the left hand side of the given expression:
$\text{LHS}=\cos\text{x}\cos\frac{\text{x}}{2}-\cos{3\text{x}}\cos\frac{9\text{x}}{2}$
We know that $2\cos\text{A}\cos\text{B}=\cos\text{A+B}+\cos(\text{A}-\text{B})$
Thus,
$\text{LHS}=\frac{1}{2}\Big[\cos\Big(\text{x}+\frac{\text{x}}{2}\Big)+\cos\Big(\text{x}-\frac{\text{x}}{2}\Big)-\frac{1}{2}\Big[\cos\Big(3\text{x}+\frac{9\text{x}}{2}\Big)+\cos\Big(3\text{x}-\frac{9\text{x}}{2}\Big)\Big]$
$=\ \frac{1}{2}\Big[\cos\Big(\frac{3\text{x}}{2}\Big)+\cos\Big(\frac{\text{x}}{2}\Big)\Big]-\frac{1}{2}\Big[\cos\Big(\frac{15\text{x}}{2}\Big)+\cos\Big(-\frac{3\text{x}}{2}\Big)\Big]$
$=\ \frac{1}{2}\Big[\cos\Big(\frac{3\text{x}}{2}\Big)-\cos\Big(\frac{\text{x}}{2}\Big)\Big]-\frac{1}{2}\Big[\cos\Big(\frac{15\text{x}}{2}\Big)+\cos\Big(-\frac{3\text{x}}{2}\Big)\Big]$$[\because\ \cos(-\theta)=\cos\theta]$
$=\ \frac{1}{2}\Big[\cos\Big(\frac{3\text{x}}{2}\Big)+\cos\Big(\frac{\text{x}}{2}\Big)-\cos\Big(\frac{15\text{x}}{2}\Big)-\cos\Big(\frac{3\text{x}}{2}\Big)\Big]$
$=\ \frac{1}{2}\Big[\cos\Big(\frac{\text{x}}{2}\Big)-\cos\Big(\frac{15\text{x}}{2}\Big)\Big]$
Also we know that,
$\cos\text{D}-\cos\text{C}=2\sin\frac{\text{C+D}}{2}\sin\frac{\text{C}-\text{D}}{2}$
Therefore,
$\text{LHS}=\frac{1}{2}\times2\sin\frac{\frac{15\text{x}}{2}+\frac{\text{x}}{2}}{2}\sin\frac{\frac{15\text{x}}{2}-\frac{\text{x}}{2}}{2}$
$=\ \sin\frac{\frac{16\text{x}}{2}}{2}\sin\frac{\frac{14\text{x}}{2}}{2}$
$=\ \sin\frac{8\text{x}}{2}\sin\frac{7\text{x}}{2}$
$=\ \text{RHS}$
Note: Question given in the book is incorrect.
RHS should be equal to $\sin\frac{8\text{x}}{2}\sin\frac{7\text{x}}{2}.$

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