Question

PROVE THAT:
cos 12 cos 24 cos 36 cos 48 Cos 60 cos 72 cos 84 equal = 1 / 128​

Answer 1

Answer:

$\boxed{\bf\;cos\,12^{\circ}\:cos\,24^{\circ}\:cos\,36^{\circ}\:cos\,48^{\circ}\:cos\,60^{\circ}\:cos\,72^{\circ}\:cos\,84^{\circ}=\frac{1}{128}}$

Step-by-step explanation:

Given:

$cos\,12^{\circ}\:cos\,24^{\circ}\:cos\,36^{\circ}\:cos\,48^{\circ}\:cos\,60^{\circ}\:cos\,72^{\circ}\:cos\,84^{\circ}$

Using

$\boxed{cos\,60^{\circ}=\frac{1}{2}}$

$=\frac{1}{2}[cos\,12^{\circ}\:cos\,24^{\circ}\:cos\,36^{\circ}\:cos\,48^{\circ}\:cos\,72^{\circ}\:cos\,84^{\circ}]$

$=\frac{1}{2}[cos\,48^{\circ}\:cos\,12^{\circ}\:cos\,72^{\circ}]\:[cos\,36^{\circ}\:cos\,24^{\circ}\:cos\,84^{\circ}]$

$=\frac{1}{2}[cos(60^{\circ}-12^{\circ})\:cos\,12^{\circ}\:cos(60^{\circ}+12^{\circ})][cos(60^{\circ}-24^{\circ})\:cos\,24^{\circ}\:cos(60^{\circ}+24^{\circ})]\:$

Using

$\boxed{\bf\;cos(60^{\circ}-A)\:cosA\:cos(60^{\circ}-A)=\frac{1}{4}\:cos3A}$

$=\frac{1}{2}[\frac{1}{4}cos3(12^{\circ})]\:[\frac{1}{4}cos3(24^{\circ})]$

$=\frac{1}{2}[\frac{1}{4}cos36^{\circ}]\:[\frac{1}{4}cos72^{\circ}]$

$=\frac{1}{32}[cos36^{\circ}\:sin18^{\circ}]$

Using

$\boxed{sin18^{\circ}=\frac{{\sqrt5}-1}{4}\:and\:cos36^{\circ}=\frac{{\sqrt5}+1}{4}}$

$=\frac{1}{32}[\frac{{\sqrt5}+1}{4}\:\frac{{\sqrt5}-1}{4}]$

$=\frac{1}{32}[\frac{(\sqrt5)^2-1^2}{16}]$

$=\frac{1}{32}[\frac{5-1}{16}]$

$=\frac{1}{32}[\frac{4}{16}]$

$=\frac{1}{32}[\frac{1}{4}]$

$=\frac{1}{128}$

$\implies\boxed{\bf\;cos\,12^{\circ}\:cos\,24^{\circ}\:cos\,36^{\circ}\:cos\,48^{\circ}\:cos\,60^{\circ}\:cos\,72^{\circ}\:cos\,84^{\circ}=\frac{1}{128}}$