Question

If two angles \( \alpha, \beta \) are such that \( 0<\alpha, \beta<\frac{\pi}{4} \), \( \sqrt{1 + \cos 2\alpha} = \frac{3}{\sqrt{5}} \) and \( \frac{\sqrt{1 - \cos 2\beta}}{\sqrt{1 + \cos 2\beta}} = \frac{1}{7} \), then \( (2\alpha + \beta) = \)

• A. \( \frac{\pi}{3} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{3\pi}{4} \)
• D. \( \frac{\pi}{4} \)

Hint:

Use the double angle formulas for cosine: \( 1 + \cos 2\theta = 2 \cos^2 \theta \) and \( 1 - \cos 2\theta = 2 \sin^2 \theta \). Find the values of \( \tan 2\alpha \) and \( \tan \beta \) from the given equations. Then use the formula for \( \tan(A + B) \) to find \( \tan(2\alpha + \beta) \). Finally, determine the value of \( 2\alpha + \beta \) within the given range.

Answer 1

The Correct Option is D

Given \( \sqrt{1 + \cos 2\alpha} = \frac{3}{\sqrt{5}} \).
We know that \( 1 + \cos 2\alpha = 2 \cos^2 \alpha \).
So, \( \sqrt{2 \cos^2 \alpha} = \frac{3}{\sqrt{5}} \).
Since \( 0<\alpha<\frac{\pi}{4} \), \( \cos \alpha>0 \), so \( \sqrt{2} |\cos \alpha| = \sqrt{2} \cos \alpha = \frac{3}{\sqrt{5}} \).
$$ \cos \alpha = \frac{3}{\sqrt{10}} $$ $$ \cos 2\alpha = 2 \cos^2 \alpha - 1 = 2 \left( \frac{9}{10} \right) - 1 = \frac{18}{10} - 1 = \frac{8}{10} = \frac{4}{5} $$ Since \( \cos 2\alpha = \frac{4}{5}>0 \) and \( 0<\alpha<\frac{\pi}{4} \), we have \( 0<2\alpha<\frac{\pi}{2} \).
Therefore, \( \sin 2\alpha = \sqrt{1 - \cos^2 2\alpha} = \sqrt{1 - \left( \frac{4}{5} \right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \).
So, \( \tan 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{3/5}{4/5} = \frac{3}{4} \).
Given \( \frac{\sqrt{1 - \cos 2\beta}}{\sqrt{1 + \cos 2\beta}} = \frac{1}{7} \).
We know that \( 1 - \cos 2\beta = 2 \sin^2 \beta \) and \( 1 + \cos 2\beta = 2 \cos^2 \beta \).
So, \( \frac{\sqrt{2 \sin^2 \beta}}{\sqrt{2 \cos^2 \beta}} = \frac{|\sin \beta|}{|\cos \beta|} = |\tan \beta| = \frac{1}{7} \).
Since \( 0<\beta<\frac{\pi}{4} \), \( \tan \beta>0 \), so \( \tan \beta = \frac{1}{7} \).
Now we need to find \( \tan(2\alpha + \beta) \).
$$ \tan(2\alpha + \beta) = \frac{\tan 2\alpha + \tan \beta}{1 - \tan 2\alpha \tan \beta} = \frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \cdot \frac{1}{7}} = \frac{\frac{21 + 4}{28}}{1 - \frac{3}{28}} = \frac{\frac{25}{28}}{\frac{25}{28}} = 1 $$ Since \( 0<2\alpha<\frac{\pi}{2} \) and \( 0<\beta<\frac{\pi}{4} \), we have \( 0<2\alpha + \beta<\frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} \).
Since \( \tan(2\alpha + \beta) = 1 \) and \( 0<2\alpha + \beta<\frac{3\pi}{4} \), we must have \( 2\alpha + \beta = \frac{\pi}{4} \).