If $\sin (\theta-\alpha), \sin \theta$ and $\sin (\theta+\alpha)$ are in H.P., then the value of $\cos 2…
If $\sin (\theta-\alpha), \sin \theta$ and $\sin (\theta+\alpha)$ are in H.P., then the value of $\cos 2 \theta$ is
- $1+4 \cos ^2 \frac{\alpha}{2}$
- $1-4 \cos ^2 \frac{\alpha}{2}$
- $-1-4 \cos ^2 \frac{\alpha}{2}$
- $-1+4 \cos ^2 \frac{\alpha}{2}$
Solution
$\sin (\theta-\alpha), \sin \theta$ 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.
$\therefore \quad \frac{2}{\sin \theta}=\frac{1}{\sin (\theta-\alpha)}+\frac{1}{\sin (\theta+\alpha)}$
$\begin{aligned} & \Rightarrow \frac{2}{\sin \theta}=\frac{\sin (\theta+\alpha)+\sin (\theta-\alpha)}{\sin (\theta-\alpha) \sin (\theta+\alpha)} \\ & \Rightarrow \frac{2}{\sin \theta}=\frac{2 \sin \theta \cos \alpha}{\sin ^2 \theta-\sin ^2 \alpha} \\ & \Rightarrow \sin ^2 \theta-\sin ^2 \alpha=\sin ^2 \theta \cos \alpha \\ & \Rightarrow \sin ^2 \theta(1-\cos \alpha)=\sin ^2 \alpha \\ & \Rightarrow \sin ^2 \theta\left(2 \sin ^2 \frac{\alpha}{2}\right)=4 \sin ^2 \frac{\alpha}{2} \cos ^2 \frac{\alpha}{2} \\ & \Rightarrow 1-\cos ^2 \theta=2 \cos ^2 \frac{\alpha}{2} \\ & \Rightarrow \cos ^2 \theta=1-2 \cos ^2 \frac{\alpha}{2} \\ & \Rightarrow 2 \cos ^2 \theta-1=1-4 \cos ^2 \frac{\alpha}{2} \\ & \Rightarrow \cos 2 \theta=1-4 \cos ^2 \frac{\alpha}{2}\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 1)
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