Question

If A = ${30}^{\circ }$, find the value of $4\cos A.\cos ({60}^{\circ }-A).\cos ({60}^{\circ }+A)$

Answer 1

Given:$A={30}^{\circ }$
$4\cos A\cos ({60}^0-A)\cos ({60}^0+A)$
= $4\cos 30\cos ({60}^0-{30}^0)\cos ({60}^0+{30}^0)$
= $4\cos {30}^0\cos {30}^0\cos {90}^0$
= $4\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\left(0\right)$
= $0$