Q184 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
Let $p = |\sin\alpha - \sin(\alpha - 90°)|$.
What is the maximum value of $p$?
Let $p = |\sin\alpha - \sin(\alpha - 90°)|$.
What is the maximum value of $p$?
- A$1$
- Correct answerB$\sqrt{2}$
- C$\sqrt{3}$
- D$2$
Solution
Given $p = |\sin\alpha - \sin(\alpha - 90^\circ)|$
Using the trigonometric identity $\sin(\alpha - 90^\circ) = -\sin(90^\circ - \alpha) = -\cos\alpha$
Substituting this into the expression for $p$
$p = |\sin\alpha - (-\cos\alpha)| = |\sin\alpha + \cos\alpha|$
The maximum value of the expression $a\sin\alpha + b\cos\alpha$ is $\sqrt{a^2 + b^2}$
Here, $a = 1$ and $b = 1$
Maximum value of $\sin\alpha + \cos\alpha = \sqrt{1^2 + 1^2} = \sqrt{2}$
Therefore, the maximum value of $p$ is $\sqrt{2}$
Answer: $\sqrt{2}$
Using the trigonometric identity $\sin(\alpha - 90^\circ) = -\sin(90^\circ - \alpha) = -\cos\alpha$
Substituting this into the expression for $p$
$p = |\sin\alpha - (-\cos\alpha)| = |\sin\alpha + \cos\alpha|$
The maximum value of the expression $a\sin\alpha + b\cos\alpha$ is $\sqrt{a^2 + b^2}$
Here, $a = 1$ and $b = 1$
Maximum value of $\sin\alpha + \cos\alpha = \sqrt{1^2 + 1^2} = \sqrt{2}$
Therefore, the maximum value of $p$ is $\sqrt{2}$
Answer: $\sqrt{2}$
Hence, the correct answer is option: 'B'.