Question

If ${\sin }^2(\theta -\alpha )\cos \alpha ={\cos }^2(\theta -\alpha )\sin \alpha =m\sin \alpha \cos \alpha$, then $\left|m\right|\ge \frac{1}{\sqrt{a}}$
Find a

Answer 1

${\sin }^2(\theta -\alpha )\cos \alpha ={\cos }^2(\theta -\alpha )\sin \alpha =m\sin \alpha \cos \alpha$
or ${\sin }^2(\theta -\alpha )=m\sin \alpha$
${\cos }^2(\theta -\alpha )=m\cos \alpha$
Adding, we get
$1=m(\sin \alpha +\cos \alpha )$
or $\sin \alpha +\cos \alpha =\frac{1}{m}$
or $\sin (\alpha +\frac{\pi }{4})=\frac{1}{\sqrt{2}m}$
or $\left|\frac{1}{m\sqrt{2}}\right|\le 1$ or $\left|m\right|\ge \frac{1}{\sqrt{2}}$
Ans: 2