If \tan x=\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}, \frac{\pi}{4} \lt \theta \lt \frac{\pi}{2}, then what is \sqrt{2}\sin x equal to?

  1. A. \sin \theta+\cos \theta ✓
  2. B. \sin \theta-\cos \theta
  3. C. \frac{\sin \theta+\cos \theta}{2}
  4. D. \frac{\sin \theta-\cos \theta}{2}

Correct Answer: A. \sin \theta+\cos \theta

Explanation

From \tan x = \frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}, we can construct a right triangle with perpendicular = \sin\theta+\cos\theta and base = \sin\theta-\cos\theta. Hypotenuse = \sqrt{(\sin\theta+\cos\theta)^2 + (\sin\theta-\cos\theta)^2} = \sqrt{2(\sin^2\theta+\cos^2\theta)} = \sqrt{2}. Thus, \sin x = \frac{\sin\theta+\cos\theta}{\sqrt{2}}, which means \sqrt{2}\sin x = \sin\theta+\cos\theta.

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