Question

If the point \( P(\sin\alpha, \cos\alpha) \) lies inside the triangle formed by the vertices \( (0, 0), \left(\frac{\sqrt{3}}{2}, 0\right), (0, \frac{\sqrt{3}}{2}) \), then \( \alpha \) lies in the interval:

• A. \( \left(0, \frac{\pi}{3}\right) \)
• B. \( \left(0, \frac{\pi}{4}\right) \)
• C. \( \left(0, \frac{\pi}{6}\right) \)
• D. \( \left(0, \frac{\pi}{12}\right) \)

Hint:

Triangle and Trigonometric Coordinates}
Plot triangle to visually inspect region
Use inequality bounds on sine and cosine
Convert coordinate constraints back to angle bounds

Answer 1

The Correct Option is D

Point \( P = (\sin\alpha, \cos\alpha) \) We are told \( P \) lies inside the triangle formed by: - \( (0,0) \) - \( \left( \frac{\sqrt{3}}{2}, 0 \right) \) - \( \left( 0, \frac{\sqrt{3}}{2} \right) \) So \( P \) must satisfy: \[ 0<\sin\alpha<\frac{\sqrt{3}}{2}, \quad 0<\cos\alpha<\frac{\sqrt{3}}{2} \] Now, both sine and cosine are decreasing and increasing respectively in the interval \( (0, \frac{\pi}{2}) \) The point at which both \( \sin\alpha = \cos\alpha = \frac{\sqrt{3}}{2} \) is not possible simultaneously, but we want both to be \(<\frac{\sqrt{3}}{2} \) Therefore, smallest angle for which both are below that value is: \[ \alpha \in \left( 0, \frac{\pi}{12} \right) \]