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Transformation Formulae — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae5 Marks
Question
Prove that:
$\cos20^\circ\cos40^\circ\cos80^\circ=\frac{1}{8}$
✓

Answer

$\text{LHS}=\cos20^\circ\cos40^\circ\cos80^\circ$
$=\ \frac{1}{2}(2\cos20^\circ\cos40^\circ)\cos80^\circ$
$=\ \frac{1}{2}[\cos(40^\circ+20^\circ)+\cos(40^\circ-20^\circ)]\cos80^\circ$$[\because\ 2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$
$=\ \frac{1}{2}[\cos60^\circ+\cos20^\circ]\cos80^\circ$
$=\ \frac{1}{2}\Big[\frac{1}{2}+\cos20^\circ\Big]\cos80^\circ$
$=\ \frac{1}{2}[\cos80^\circ+2\cos20^\circ\cos80^\circ]$
$=\ \frac{1}{4}[\cos80^\circ+\cos(80^\circ+20^\circ)+\cos(20^\circ-80^\circ)]$
$=\ \frac{1}{4}[\cos80^\circ+\cos100^\circ+\cos60^\circ]$
$=\ \frac{1}{4}[\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ]$
$=\ \frac{1}{4}[\cos80^\circ-\cos80^\circ+\cos60^\circ]$
$= \frac{1}{4}\Big[\frac{1}{2}\Big]=\frac{1}{8}=\text{RHS}$

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