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

$\begin{array}{l}\\ \text{We have}{m}^2+n^2=[{\left(a\cos \theta +b\sin \theta \right)}^2+{\left(a\sin \theta -b\cos \theta \right)}^2]\\ \text{= (}{a}^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\cos \theta \sin \theta )+(a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\sin \theta \cos \theta )\\ \text{=}{a}^2{\cos }^2\theta +b^2{\sin }^2\theta +a^2{\sin }^2\theta +b^2{\cos }^2\theta \\ \text{=(}{a}^2{\cos }^2\theta +a^2{\sin }^2\theta )+(b^2{\cos }^2\theta +b^2{\sin }^2\theta )\\ =a^2({\cos }^2\theta +{\sin }^2\theta )+b^2({\cos }^2\theta +{\sin }^2\theta )\\ =a^2+b^2\text{[}\because {\text{sin}}^2+{\cos }^2=1]\\ {\text{Hence, m}}^2+n^2=a^2+b^2\end{array}$