Question

Prove that $4\cos {12}^o\cos {12}^o\cos {48}^o\cos {72}^o=\cos {36}^o$.

Answer 1

$4\cos {12}^o\cos {48}^o\cos {72}^o$

$=2\left(2\cos {12}^o\cos {48}^o\right)\cos {72}^o$

$=2\left[\cos \right(12+48)+\cos (12-48\left)\right]\cos {72}^o$

$=2[\cos {60}^o+\cos {36}^o]\cos {72}^o$

$=2\cos {60}^o\cos {72}^o+2\cos {36}^o\cos {72}^o$

$=2\times \frac{1}{2}\cos {72}^o+\left[\right(\cos 36+72)+\cos (36-7\left)\right]$

$=\cos {72}^o+\cos {108}^o+\cos {36}^o$

$=2\cos ({72}^o+{108}^o)/2\cos ({72}^o-{108}^o)/2+\cos {36}^o$

$=2\cos {90}^o\cos {18}^o+\cos {36}^o$

$=\cos {36}^o$ $[\because \cos {90}^o=0]$.