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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Prove that:
$\cos40^\circ\cos80^\circ\cos160=-\frac{1}{8}$
✓

Answer

$\cos40^\circ\cos80^\circ\cos160^\circ=-\frac{1}{8}$
$\text{LHS}=\cos40^\circ\cos80^\circ\cos160^\circ$
$=\ \cos80^\circ\cos40^\circ\cos160^\circ$
Multiplying and dividing by 2
$=\ \frac{1}{2}(\cos80^\circ\times(2\cos40^\circ\cos160^\circ))$
$2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})$
$=\ \frac{1}{2}(\cos80^\circ(\cos(40^\circ+160^\circ)+\cos(40^\circ-160^\circ)))$
$=\ \frac{1}{2}(\cos80^\circ(\cos200+\cos(-120)))$
$=\ \frac{1}{2}(\cos80^\circ(\cos180^\circ+20^\circ)+\cos(180^\circ-60^\circ))$
$=\ \frac{1}{2}\cos80^\circ(\cos20^\circ+\cos60^\circ)$
$=\ \frac{1}{2}\cos80^\circ\cos20^\circ+\frac{1}{2}\cos80^\circ+60^\circ$
$=\ -\frac{1}{2}(2\cos80^\circ\cos20^\circ)+\frac{1}{2}\cos80^\circ\cos60^\circ$
$=\ -\frac{1}{4}[2\cos80^\circ\cos20^\circ+\cos80^\circ]$
$=\ -\frac{1}{4}[\cos(80^\circ+20^\circ)+\cos(80^\circ-20^\circ)+\cos80^\circ]$
$=\ -\frac{1}{4}[\cos100^\circ+\cos60^\circ+\cos80^\circ]$
$=\ -\frac{1}{4}[\cos(180^\circ-80^\circ)+\cos60^\circ+\cos80^\circ]$
$=\ -\frac{1}{4}[-\cos80^\circ+\cos60^\circ+\cos80^\circ]$
$=\ -\frac{1}{4}\cos60^\circ$
$=\ -\frac{1}{4}\times\frac{1}{2}$
$=\ -\frac{1}{8}\ \text{RHS}$

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