Question

Let \( \theta \in [0, \frac{\pi}{2}] \). Which one of the following is true?

• A. \( \sin^2 \theta > \cos^2 \theta \)
• B. \( \sin^2 \theta < \cos^2 \theta \)
• C. \( \sin \theta > \cos \theta \)
• D. \( \cos \theta > \sin \theta \)
• E. \( \sin \theta + \cos \theta \le \sqrt{2} \)

Hint:

The expression \( \sin\theta + \cos\theta \) reaches its maximum value of \( \sqrt{2} \) exactly at \( \theta = 45^\circ \) (or \( \pi/4 \)).

Answer 1

Answer: (E)

Concept: The maximum value of the expression \( a\sin\theta + b\cos\theta \) is given by \( \sqrt{a^2 + b^2} \).

Step 1: Find the maximum value of \( \sin \theta + \cos \theta \).
Here, \( a = 1 \) and \( b = 1 \). The maximum value is: \[ \sqrt{1^2 + 1^2} = \sqrt{2} \]

Step 2: Verify the range.
For any \( \theta \), \( -\sqrt{2} \le \sin\theta + \cos\theta \le \sqrt{2} \). Since \( \theta \in [0, \frac{\pi}{2}] \), both \( \sin\theta \) and \( \cos\theta \) are non-negative, so the sum is between 1 and \( \sqrt{2} \). Thus, the statement \( \sin\theta + \cos\theta \le \sqrt{2} \) is universally true for the given interval.