By signing in you confirm you are 18 or older, or have your parent’s or guardian’s consent.

NDA 1 2025 Maths, question 34

Topic: Trigonometry

Passage · Q34 to Q35

Let p=∣sin⁡α−sin⁡(α−90∘)∣p = \left| \sin\alpha - \sin\left( \alpha - 90^{\circ} \right) \right|.

Q34

What is the minimum value of pp?

  1. (a) 0Correct answer
  2. (b) 1/21/2
  3. (c) 1/21/\sqrt{2}
  4. (d) 1

Solution.

Answer: (a) 0

  1. Since sin⁡(α−90∘)=−cos⁡α\sin\left( \alpha - 90^{\circ} \right) = - \cos\alpha, we have p=∣sin⁡α+cos⁡α∣=2∣sin⁡(α+45∘)∣p = \left| \sin\alpha + \cos\alpha \right| = \sqrt{2}\left| \sin\left( \alpha + 45^{\circ} \right) \right|.
  2. An absolute sine can be zero, for example at α=135∘\alpha = 135^{\circ}.
  3. Therefore the minimum is 0.