If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ - y cos θ = 0, for every \(\theta \in\left(0, \frac{\pi}{2}\right) \), then what is x2 + y2 equal to ?

Given:

For every \(θ \in\left(0, \frac{\pi}{2}\right) \)

x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ - y cos θ = 0

Formula Used:

1. (a + b)² = a² + b² + 2ab

2. (a - b)² = a² + b² - 2ab

Calculation:

Since every \(θ \in\left(0, \frac{\pi}{2}\right) \), putting θ = 45⁰

According to the question

⇒ x sin3 θ + y cos3 θ = sin θ cos θ

⇒ x \((\frac{1}{√2})^3\) + y \((\frac{1}{√2})^3\) = \(\frac{1}{√2}\) × \(\frac{1}{√2}\)

⇒ \(\frac{x}{2√2}\) + \(\frac{y}{2√2}\) = \(\frac{1}{2}\)

⇒ x + y = √2     -----(1)

Again, according to the question

⇒ x sin θ - y cos θ = 0

⇒ x \((\frac{1}{√2})\) - y \((\frac{1}{√2})\) = 0

⇒ x - y = 0        -----(2)

Squaring Equation(1) and (2) and adding

⇒ x² + y² + 2xy + x² + y² - 2xy = (√2)² + 0

⇒ 2(x² + y²) = 2

⇒ x² + y² = 1

∴ The value of x² + y² is 1.