Question

Find $a$ if, $\cos {36}^0\cos {72}^0\cos {108}^0\cos {144}^0=\frac{1}{2a}$

Answer 1

We have, $\cos {36}^0\cos {72}^0\cos {108}^0\cos {144}^0$

$=\cos {36}^0\sin {18}^0(-\sin {18}^0)(-\cos {36}^0)$

$={\cos }^2{36}^0{\sin }^2{18}^0={\left(\frac{\sqrt{5}+1}{4}\right)}^2{\left(\frac{\sqrt{5}-1}{4}\right)}^2$

$={\left[\left(\frac{\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}-1}{4}\right)\right]}^2=\frac{1}{16}$
Ans:a=8