Question:
If \(\cos\alpha+\cos\beta=\frac{3}{2}\) and \(\sin\alpha+\sin\beta=\frac{1}{2}\) and \(\theta\) is the arithmetic mean of \(\alpha\) and \(\beta\), then \(\sin2\theta+\cos2\theta\) is equal to :
If \(\cos\alpha+\cos\beta=\frac{3}{2}\) and \(\sin\alpha+\sin\beta=\frac{1}{2}\) and \(\theta\) is the arithmetic mean of \(\alpha\) and \(\beta\), then \(\sin2\theta+\cos2\theta\) is equal to :
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Use sum-to-product, divide to get tan(theta) = 1/3, then express sin 2theta and cos 2theta in tan theta.
Updated On: Nov 11, 2026
- \(\frac{3}{5}\)
- \(\frac{4}{5}\)
- \(\frac{7}{5}\)
- \(\frac{8}{5}\)
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The Correct Option is C
Solution and Explanation
Step 1: \(\theta=\frac{\alpha+\beta}{2}\). \(\cos\alpha+\cos\beta=2\cos\theta\cos\frac{\alpha-\beta}{2}=\frac{3}{2}\) and \(\sin\alpha+\sin\beta=2\sin\theta\cos\frac{\alpha-\beta}{2}=\frac{1}{2}\).
Step 2: Dividing, \(\tan\theta=\frac{1}{3}\).
Step 3: \(\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2/3}{10/9}=\frac{3}{5}\); \(\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{8/9}{10/9}=\frac{4}{5}\).
Step 4: \(\sin2\theta+\cos2\theta=\frac{3}{5}+\frac{4}{5}=\frac{7}{5}\).
Answer: (3) \(\frac{7}{5}\)
Step 2: Dividing, \(\tan\theta=\frac{1}{3}\).
Step 3: \(\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2/3}{10/9}=\frac{3}{5}\); \(\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{8/9}{10/9}=\frac{4}{5}\).
Step 4: \(\sin2\theta+\cos2\theta=\frac{3}{5}+\frac{4}{5}=\frac{7}{5}\).
Answer: (3) \(\frac{7}{5}\)
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