Question

If a cos θ – b sin θ = m and a sin θ + b cos θ = n, prove that: a² + b² = m² + n²

Answer 1

m = a cos θ – b sin θ   ...(1)

n = a sin θ + b cos θ   ...(2)

Square the (1) equation:

m² = (a cos θ – b sin θ)²

Using the identity: (x − y)² = x² + y² − 2xy

m² = a² cos² θ + b² sin² θ − 2ab sin θ . cos θ    ...(3)

Square the (2) equation:

n² = (a sin θ + b cos θ)²

Using the identity: (x + y)² = x² + y² + 2xy

n² = a² sin² θ + b² cos² θ + 2ab sin θ . cos θ   ...(4)

Add Equation 3 and Equation 4

m² + n² = (a² cos² θ + b² sin² θ − 2ab sin θ . cos θ) + (a² sin² θ + b² cos² θ + 2ab sin θ . cos θ)

Simplify the expression:

m² + n² = a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ

m² + n² = a²(cos² θ + sin² θ) + b²(sin² θ + cos² θ)

Apply the Trigonometric Identity:

m² + n² = a²(1) + b²(1)

m² + n² = a² + b²