Question:
If \(P(\sin\alpha,\cos\alpha)\) lies inside the triangle formed by the vertices
\[
(0,0),\left(\frac{\sqrt3}{2},0\right),\left(0,\frac{\sqrt3}{2}\right),
\]
then \(\alpha\) lies in the interval:
If \(P(\sin\alpha,\cos\alpha)\) lies inside the triangle formed by the vertices
\[
(0,0),\left(\frac{\sqrt3}{2},0\right),\left(0,\frac{\sqrt3}{2}\right),
\]
then \(\alpha\) lies in the interval:
Show Hint
Whenever \(\sin\alpha+\cos\alpha\) appears, convert it into
\[
\sqrt2\sin\left(\alpha+\frac{\pi}{4}\right)
\]
to simplify the inequality.
Updated On: Jun 26, 2026
- \(\left(0,\frac{\pi}{3}\right)\)
- \(\left(0,\frac{\pi}{4}\right)\)
- \(\left(0,\frac{\pi}{6}\right)\)
- \(\left(0,\frac{\pi}{12}\right)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Find the equation of the hypotenuse of the triangle.
The vertices are \[ (0,0),\quad \left(\frac{\sqrt3}{2},0\right), \quad \left(0,\frac{\sqrt3}{2}\right). \] Hence the hypotenuse is \[ x+y=\frac{\sqrt3}{2}. \] The interior of the triangle satisfies \[ x+y\lt \frac{\sqrt3}{2}. \]
Step 2: Substitute the coordinates of \(P\).
Since \[ P=(\sin\alpha,\cos\alpha), \] for \(P\) to lie inside the triangle, \[ \sin\alpha+\cos\alpha \lt \frac{\sqrt3}{2}. \] Using \[ \sin\alpha+\cos\alpha = \sqrt2\sin\left(\alpha+\frac{\pi}{4}\right), \] we get \[ \sqrt2\sin\left(\alpha+\frac{\pi}{4}\right) \lt \frac{\sqrt3}{2}. \]
Step 3: Solve the inequality.
This gives \[ \sin\left(\alpha+\frac{\pi}{4}\right) \lt \frac{\sqrt3}{2\sqrt2}. \] On solving and using the admissible range for \(\alpha\), we obtain \[ 0\lt \alpha\lt \frac{\pi}{12}. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\alpha\in\left(0,\frac{\pi}{12}\right)}. \]
The vertices are \[ (0,0),\quad \left(\frac{\sqrt3}{2},0\right), \quad \left(0,\frac{\sqrt3}{2}\right). \] Hence the hypotenuse is \[ x+y=\frac{\sqrt3}{2}. \] The interior of the triangle satisfies \[ x+y\lt \frac{\sqrt3}{2}. \]
Step 2: Substitute the coordinates of \(P\).
Since \[ P=(\sin\alpha,\cos\alpha), \] for \(P\) to lie inside the triangle, \[ \sin\alpha+\cos\alpha \lt \frac{\sqrt3}{2}. \] Using \[ \sin\alpha+\cos\alpha = \sqrt2\sin\left(\alpha+\frac{\pi}{4}\right), \] we get \[ \sqrt2\sin\left(\alpha+\frac{\pi}{4}\right) \lt \frac{\sqrt3}{2}. \]
Step 3: Solve the inequality.
This gives \[ \sin\left(\alpha+\frac{\pi}{4}\right) \lt \frac{\sqrt3}{2\sqrt2}. \] On solving and using the admissible range for \(\alpha\), we obtain \[ 0\lt \alpha\lt \frac{\pi}{12}. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\alpha\in\left(0,\frac{\pi}{12}\right)}. \]
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