Question

Condition that should be imposed on a and b such that $S_2$ is non-empty -

• A. $\left|\frac{a}{2}\sin b\right|<1$
• B. $\left|\frac{a}{2}\sin b\right|\le 1$
• C. $|a\sin b|\le 1$
• D. none of these

Answer 1

Answer: (C)

$|a\sin b|\le 1$

$(1+a)\cos \theta \cos (2\theta -b)=(1+a\cos 2\theta )\cos (\theta -b),\Rightarrow \cos \theta \cos (2\theta -b)+a\cos \theta \cos (2\theta -b)=\cos (\theta -b)+a\cos 2\theta \cos (\theta -b),\Rightarrow 2\cos \theta \cos (2\theta -b)+2a\cos \theta \cos (2\theta -b)=2\cos (\theta -b)+2a\cos 2\theta \cos (\theta -b),\Rightarrow \cos (3\theta -b)+\cos (\theta -b)+a\left(\cos \right(3\theta -b)+\cos (\theta -b\left)\right)=2\cos (\theta -b)+a\left(\cos \right(3\theta -b)+\cos (\theta +b\left)\right),\Rightarrow \cos (3\theta -b)+a\cos (\theta -b)=\cos (\theta -b)+a\cos (\theta +b),\Rightarrow \cos (3\theta -b)-\cos (\theta -b)=a\left(\cos \right(\theta +b)-\cos (\theta -b\left)\right),\Rightarrow 2\sin (2\theta -b)\sin \theta =2a\sin \theta \sin b\Rightarrow \sin \theta \left(\sin \right(2\theta -b)-a\sin b)=0,\Rightarrow \sin \theta =0or\sin (2\theta -b)=a\sin b,\Rightarrow \sin \theta =0\Rightarrow \theta =n\pi ,nϵZand\sin (2\theta -b)=a\sin b.\Rightarrow \sin (2\theta -b)=\sin \left({\sin }^{-1}\right(a\sin b\left)\right),\Rightarrow 2\theta -b=n\pi +\left(-1{)}^n{\sin }^{-1}\right(a\sin b)nϵZ,|a\sin b|\le 1\to for{S}_2=nonempty$