Question

If $\cos \theta, \sqrt{2}\sin \theta$ and $\sqrt{3}\tan \theta$, where $0 < \theta < \frac{\pi}{2}$, are the second, third and fourth terms of a geometric series, respectively, then the first term of the geometric series is

• A. $\frac{1}{2\sqrt{6}}$
• B. $\frac{3}{2\sqrt{6}}$
• C. $\frac{1}{\sqrt{6}}$
• D. $\frac{3}{\sqrt{6}}$
• E. $\frac{2}{\sqrt{6}}$

Hint:

In GP problems involving trigonometry, equate the common ratio $r = T_n / T_{n-1}$ to find the angle first. This usually simplifies the radical expressions significantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For three consecutive terms \(x, y, z\) in a geometric progression, the middle term is the geometric mean of the other two: \(y^2 = xz\).
We use this to find \(\theta\).
Step 2: Key Formula or Approach:
Given \(a_2 = \cos \theta\), \(a_3 = \sqrt{2}\sin \theta\), \(a_4 = \sqrt{3}\tan \theta\).
Use the property: \((a_3)^2 = a_2 \cdot a_4\).
Once \(\theta\) is found, the common ratio \(r\) is \(\frac{a_3}{a_2}\), and the first term \(a_1 = \frac{a_2}{r}\).
Step 3: Detailed Explanation:
Set up the geometric mean equation:
\[ (\sqrt{2}\sin \theta)^2 = (\cos \theta)(\sqrt{3}\tan \theta) \] \[ 2\sin^2 \theta = \sqrt{3} \cos \theta \left(\frac{\sin \theta}{\cos \theta}\right) \] \[ 2\sin^2 \theta = \sqrt{3}\sin \theta \] Since \(0<\theta<\frac{\pi}{2}\), \(\sin \theta \neq 0\). We can divide by \(\sin \theta\):
\[ 2\sin \theta = \sqrt{3} \implies \sin \theta = \frac{\sqrt{3}}{2} \] Thus, \(\theta = \frac{\pi}{3}\).
Now find the terms:
\(a_2 = \cos(\frac{\pi}{3}) = \frac{1}{2}\).
\(a_3 = \sqrt{2}\sin(\frac{\pi}{3}) = \sqrt{2}\frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2}\).
Find the common ratio \(r\):
\[ r = \frac{a_3}{a_2} = \frac{\frac{\sqrt{6}}{2}}{\frac{1}{2}} = \sqrt{6} \] Find the first term \(a_1\):
\[ a_1 = \frac{a_2}{r} = \frac{\frac{1}{2}}{\sqrt{6}} = \frac{1}{2\sqrt{6}} \] Step 4: Final Answer:
The first term is \(\frac{1}{2\sqrt{6}}\).