QQuestion
If 0° < θ < 90° then sin θ < sin²θ
If 0° < θ < 90° then sin θ < sin²θ
False
Explanation:
We know that for 0° < θ < 90°,
0 < sin θ < 1
Since squaring a fraction between 0 and 1 makes it smaller, we get:
sin² θ < sin θ
But the given statement claims sin θ < sin²θ, which is incorrect.
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