Question

$A=\cos {20}^0\cos {40}^0\cos {60}^0\cos {80}^0$
$B=\cos 6^0\cos {42}^0\cos {66}^0\cos {78}^0$
$C=\cos {36}^0\cos {72}^{0}cos{108}^{0}cos{144}^0$

• A. $A>B>C$
• B. $B>C>A$
• C. $C>A>B$
• D. $A=B=C$

Answer 1

The correct option is D $A=B=C$

$A=\cos 20.\cos 40.\cos 60.\cos 80=\frac{2sin20.\cos 20.\cos 40.\cos 60.\cos 80}{2sin20}=\frac{2sin40.\cos 40.\cos 60.\cos 80}{4sin20}=\frac{2sin80.\cos 80.\cos 60}{8sin20}=\frac{sin160}{16sin20}=\frac{sin(180-20)}{16sin20}=\frac{1}{16}B=\cos 6.\cos 42.\cos 66.\cos 78=\frac{1}{4}\left(2\cos 6.\cos 66\right)\left(2\cos 42.\cos 78\right)=\frac{1}{4}(\cos 72+\cos 60)(\cos 120+\cos 36)=\frac{1}{4}(\cos 72+\frac{1}{2})(\cos 36-\frac{1}{2})=\frac{1}{4}(\frac{\sqrt{5}-1}{4}+\frac{1}{2})(\frac{\sqrt{5}+1}{4}-\frac{1}{2})=\frac{1}{16}C=\cos 36.\cos 72.\cos 108.\cos .144=\cos 36.\cos 72.\cos (180-72).\cos (180-36)=\cos 36.\cos 72.\cos 72.\cos 36={\left(\cos 36.\cos 72\right)}^2={(\frac{\sqrt{5}-1}{4}.\frac{\sqrt{5}+1}{4})}^2={\left(\frac{1}{4}\right)}^2=\frac{1}{16}$
Hence, $A=B=C$