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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Prove that:
$4\cos\text{x}\cos\Big(\frac{\pi}{3}+\text{x}\Big)\cos\Big({\frac{\pi}{3}-\text{x}}\Big)=\cos3\text{x}$
✓

Answer

We have,
$\text{LHS}=4\cos\text{x}\cos\Big(\frac{\pi}{3}+\text{x}\Big)\cos\Big(\frac{\pi}{3}-\text{x}\Big)$
$=\ 2\cos\text{x}\Big[2\cos\Big(\frac{\pi}{3}+\text{x}\Big)\cos\Big(\frac{\pi}{3}-\text{x}\Big)\Big]$
$=\ 2\cos\text{x}\Big[2\cos\Big(\frac{\pi}{3}+\text{x}+\frac{\pi}{3}-\text{x}\Big)+\cos\Big(\frac{\pi}{3}+\text{x}-\frac{\pi}{3}+\text{x}\Big)\Big]$
$=\ 2\cos\text{x}\Big[\cos\frac{2\pi}{3}+\cos2\text{x}\Big]$
$=\ 2\cos\text{x}\Big[\cos\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)+\cos2\text{x}\Big]$
$=\ 2\cos\text{x}\Big[-\sin\frac{\pi}{6}+\cos2\text{x}\Big]$
$=\ 2\cos\text{x}\Big[-\frac{1}{2}+\cos2\text{x}\Big]$
$=\ -2\cos\text{x}\times\frac{1}{2}+2\cos\text{x}\cos2\text{x}$
$=\ -\cos\text{x}+[\cos(\text{x}+2\text{x})+\cos(2\text{x}-\text{x})]$
$=\ -\cos\text{x}+\cos3\text{x}+\cos\text{x}$
$=\ \cos3\text{x}$
$=\ \text{RHS}$
LHS = RHS Hence Proved.

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