Question

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

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 + 2 ab cos theta sin theta) + (a^2 sin^2 theta + b^2 cos^2 theta -2ab cos theta sin 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 + b^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 + cos^2 = 1]`

Hence, `m^2 + n^2 = a^2 + b^2`