What is the <strong>MINIMUM</strong> value of p?
Consider the following for the two (02) items that follow : Let p=|\sin \alpha-\sin(\alpha-90^{\circ})|.
- A. 0 ✓
- B. 1/2
- C. 1/\sqrt{2}
- D. 1
Correct Answer: A. 0
Explanation
p = |\sin\alpha - \sin(\alpha-90^{\circ})| = |\sin\alpha - (-\cos\alpha)| = |\sin\alpha + \cos\alpha|. This can be rewritten as p = |\sqrt{2}\sin(\alpha+45^{\circ})|. The minimum value of an absolute function is 0, which occurs when \alpha = -45^{\circ}.