Question

If $A={30}^0$, then $4\cos A\cos ({60}^0-A)\cos ({60}^0+A)=\cos 3A$

If above statement is true enter 1 else 0

Answer 1

Given:$A={30}^{\circ }$
Now, $4\cos A\cos ({60}^0-A)\cos ({60}^0+A)=\cos 3A$
L.H.S: $4\cos A\cos ({60}^0-A)\cos ({60}^0+A)$
= $4\cos 30\cos ({60}^0-{30}^0)\cos ({60}^0+{30}^0)$
= $4\cos {30}^0\cos {30}^0\cos {90}^0$
= $4\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}\cdot \left(0\right)$
= $0$
R.H.S: $\cos 3A=\cos 3\times {30}^0=\cos {90}^0=0$

$4\cos A\cos ({60}^0-A)\cos ({60}^0+A)=\cos 3A$ is true for $A={30}^{\circ }$.