If \tan \alpha=\frac{1}{7}, \sin \beta=\frac{1}{\sqrt{10}}, 0 \lt \alpha, \beta \lt \frac{\pi}{2}, then what is the value of \cos(\alpha+2\beta)?
- A. -\frac{1}{2}
- B. -\frac{1}{\sqrt{2}}
- C. \frac{1}{\sqrt{2}} ✓
- D. \frac{1}{2}
Correct Answer: C. \frac{1}{\sqrt{2}}
Explanation
Since \sin\beta = \frac{1}{\sqrt{10}} in the first quadrant, \cos\beta = \frac{3}{\sqrt{10}} and \tan\beta = \frac{1}{3}. Using the double angle formula, \tan2\beta = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{8/9} = \frac{3}{4}. Then \tan(\alpha+2\beta) = \frac{\tan\alpha + \tan2\beta}{1 - \tan\alpha\tan2\beta} = \frac{1/7 + 3/4}{1 - (1/7)(3/4)} = 1. Since 0 \lt \alpha+2\beta \lt \frac{3\pi}{2}, it follows \alpha+2\beta = \frac{\pi}{4}. Thus \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}.