Question 91

Mathematics Trigonometric Equations Easy

If \( \sin \beta \) is the GM between \( \sin \alpha \) and \( \cos \alpha \) , then \( \cos 2 \beta \) is equal to

(A) \( 2 \sin ^{2}\left(\frac{3\pi}{4}-\alpha\right) \)
(B) \( 2 \sin ^{2}\left(\frac{\pi}{4}-\alpha\right) \)
(C) \( 2 \cos ^{2}\left(\frac{\pi}{4}-\alpha\right) \)
(D) \( 2 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right) \)
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

We are given that \(\sin \beta\) is the geometric mean (GM) of \(\sin \alpha\) and \(\cos \alpha\). This means: \[ \sin \beta = \sqrt{\sin \alpha \cdot \cos \alpha} \] We need to find \(\cos 2\beta\). Using the double-angle identity: \[ \cos 2\beta = 1 - 2 \sin^2 \beta \] Substituting \(\sin \beta = \sqrt{\sin \alpha \cos \alpha}\): \[ \sin^2 \beta = \sin \alpha \cos \alpha \] So, \[ \cos 2\beta = 1 - 2 \sin \alpha \cos \alpha \] Using the identity: \[ \sin \alpha \cos \alpha = \frac{1}{2} \sin 2\alpha \] we get: \[ \cos 2\beta = 1 - 2 \times \frac{1}{2} \sin 2\alpha = 1 - \sin 2\alpha \] Now, using another identity: \[ 1 - \sin 2\alpha = 2 \sin^2 \left(\frac{\pi}{4} - \alpha \right) \] Thus, \[ \cos 2\beta = 2 \sin^2 \left(\frac{\pi}{4} - \alpha \right) \]

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