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Transformation Formulae — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae2 Marks
Question
Prove that:
$\cos\Big(\frac{3\pi}{4}+\text{x}\Big)-\cos\Big(\frac{3\pi}{4}-\text{x}\Big)=-\sqrt2\sin\text{x}$
✓

Answer

We have,
$\text{LHS}=\cos\Big(\frac{3\pi}{4}-\text{x}\Big)-\cos\Big(\frac{3\pi}{4}+\text{x}\Big)$
$=\ -\Big[\cos\Big(\frac{3\pi}{4}-\text{x}\Big)-\cos\Big(\frac{3\pi}{4}+\text{x}\Big)\Big]$
$=\ -\Big[2\sin\frac{3\pi}{4}\sin\text{x}\Big]$ $[\because\ \cos(\text{A}-\text{B})-\cos(\text{A+B})=2\sin\text{A}\sin\text{B}]$
$=\ -2\sin\Big(\frac{\pi}{2}+\frac{\pi}{4}\Big)\sin\text{x}$
$=\ -2\cos\frac{\pi}{4}\sin\text{x}$
$=\ -2\times\frac{1}{\sqrt2}\times\sin\text{x}$
$=\ -\frac{\sqrt2\times\sqrt2}{\sqrt2}\sin\text{x}$
$=\ -\sqrt2\sin\text{x}$
$=\ \text{RHS}$
$\therefore\ \cos\Big(\frac{3\pi}{4}+\text{x}\Big)-\cos\Big(\frac{3\pi}{4}-\text{x}\Big)=-\sqrt2\sin\text{x}\ \text{Hence proved.}$

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