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

We have `m^2 + n^2 = [(a cos theta + b sin theta)^2 + (a sin theta - b cos theta)^2]` 

`= (a^2 cos^2 theta + b^2 sin^2 theta + 2ab cos theta sin theta) + (a^2sin^2 theta + b^2cos^2 theta - 2ab sin theta cos theta)`

`= a^2 cos^2 theta + b^2 sin^2 theta + a^2 sin^2 theta + b^2 cos^2 theta`

`= (a^2 cos^2 theta + a^2 sin^2 theta) + (b^2 cos^2 theta + b^2 sin^2 theta)`

`= a^2(cos^2 theta + sin^2 theta) + b^2(cos^2 theta + sin^2 theta)`

`=a^2 + b^2`   [∵ `sin^2 theta + cos^2 theta = 1]`

Hence, m² + n² = a² + b²