<p>If <math>\sin^{100} \theta - \cos^{100} \theta = 1</math>, then <math>\theta</math> is</p>
<p>(a) <math>2n\pi + \frac{\pi}{3}</math>, <math>n\in\mathbb{Z}</math></p>
<p>(b) <math>n\pi + \frac{\pi}{2}</math>, <math>n\in\mathbb{Z}</math></p>
<p>(c) <math>n\pi + \frac{\pi}{2}</math>, <math>n\in\mathbb{Z}</math></p>
<p>(d) <math>2n\pi - \frac{\pi}{2}</math>, <math>n\in\mathbb{Z}</math></p>
Step-by-Step Solution
Key Concept: Since sin²θ + cos²θ = 1, at least one of |sin θ| or |cos θ| must equal 1. For sin¹⁰⁰θ - cos¹⁰⁰θ = 1 to hold, we need sin θ = ±1 and cos θ = 0 simultaneously.
<p><strong>Step 1:</strong> Note that sin²θ + cos²θ = 1, which means both |sin θ| ≤ 1 and |cos θ| ≤ 1.</p><p><strong>Step 2:</strong> Since both sin¹⁰⁰θ and cos¹⁰⁰θ are non-negative (even powers), and sin¹⁰⁰θ - cos¹⁰⁰θ = 1, we need sin¹⁰⁰θ = 1 + cos¹⁰⁰θ.</p><p><strong>Step 3:</strong> For sin¹⁰⁰θ ≤ 1, we require 1 + cos¹⁰⁰θ ≤ 1, which means cos¹⁰⁰θ ≤ 0. But cos¹⁰⁰θ ≥ 0 always, so cos¹⁰⁰θ = 0.</p><p><strong>Step 4:</strong> If cos¹⁰⁰θ = 0, then cos θ = 0. From sin¹⁰⁰θ - 0 = 1, we get sin¹⁰⁰θ = 1, so sin θ = ±1.</p><p><strong>Step 5:</strong> Since cos θ = 0 and sin θ = ±1, both conditions are satisfied when θ = π/2 + nπ for n ∈ ℤ, which can be written as nπ + π/2.</p><p><strong>Step 6:</strong> Verification: At θ = π/2: sin θ = 1, cos θ = 0, so 1¹⁰⁰ - 0¹⁰⁰ = 1 ✓. At θ = 3π/2: sin θ = -1, cos θ = 0, so (-1)¹⁰⁰ - 0¹⁰⁰ = 1 - 0 = 1 ✓</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B