If \(\sin (\theta-\alpha), \sin \theta\) and \(\sin (\theta+\alpha)\) are in H.P., then the value of \(\cos 2 \theta\) is

\(\sin (\theta-\alpha), \sin \theta\) and \(\sin (\theta+\alpha)\) are in H.P. \(\Rightarrow \frac{1}{\sin (\theta-\alpha)}, \frac{1}{\sin \theta}, \frac{1}{\sin (\theta+\alpha)}\) will be in A.P.
\(\therefore \frac{2}{\sin \theta}=\frac{1}{\sin (\theta-\alpha)}+\frac{1}{\sin (\theta+\alpha)}\)
\(\Rightarrow \frac{2}{\sin \theta}=\frac{\sin (\theta+\alpha)+\sin (\theta-\alpha)}{\sin (\theta-\alpha) \sin (\theta+\alpha)} \)
\( \Rightarrow \frac{2}{\sin \theta}=\frac{2 \sin \theta \cos \alpha}{\sin ^2 \theta-\sin ^2 \alpha} \)
\( \Rightarrow \sin ^2 \theta-\sin ^2 \alpha=\sin ^2 \theta \cos \alpha \)
\( \Rightarrow \sin ^2 \theta(1-\cos \alpha)=\sin ^2 \alpha \)
\( \Rightarrow \sin ^2 \theta\left(2 \sin ^2 \frac{\alpha}{2}\right)=4 \sin ^2 \frac{\alpha}{2} \cos ^2 \frac{\alpha}{2} \)
\( \Rightarrow 1-\cos ^2 \theta=2 \cos ^2 \frac{\alpha}{2} \)
\( \Rightarrow \cos ^2 \theta=1-2 \cos ^2 \frac{\alpha}{2} \)
\( \Rightarrow 2 \cos ^2 \theta-1=1-4 \cos ^2 \frac{\alpha}{2} \)
\( \Rightarrow \cos 2 \theta=1-4 \cos ^2 \frac{\alpha}{2}\)