Question

If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then show that a² + b² = m² + n².

Answer 1

Given,

a cos θ + b sin θ = m …(1)

a sin θ – b cos θ = n …(2)

Squaring and adding equation 1 and 2, we get –

(a cos θ + b sin θ)² + (a sin θ – b cos θ)² = m² + n²

⇒ a²cos²θ + b²sin²θ + 2ab sin θ cos θ + a²sin²θ + b²cos²θ - 2ab sin θ cos θ = m² + n²

⇒ a²cos²θ + b²sin²θ + a²sin²θ + b²cos²θ = m² + n²

⇒ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ) = m² + n²

Using: sin²θ + cos²θ = 1, we get –

⇒ a² + b² = m² + n²