Question

If x = a sinθ + bcosθ and y = acosθ – bsinθ, prove that $x^2+y^2=a^2+b^2.$

Answer 1

$$\begin{gathered} \text{Given:}x=a\sin \theta +b\cos \theta \\ \text{Squaring both sides, we get:} \\ {x}^2=a^2{\sin }^2\theta +2ab\sin \theta \cos \theta +b^2{\cos }^2\theta ...\left(\text{i}\right) \\ \text{Also,}y=a\cos \theta -b\sin \theta \\ \text{Squaring both sides, we get:} \\ {y}^2\text{=}{a}^2{\cos }^2\theta -2ab\sin \theta \cos \theta +b^2{\sin }^2\theta ...\left(\text{ii}\right) \\ \therefore \text{LHS=}{x}^2+y^2 \\ \text{=}{a}^2{\sin }^2\theta +2ab\sin \theta \cos \theta +b^2{\cos }^2\theta + \\ {a}^2{\cos }^2\theta -2ab\sin \theta \cos \theta +b^2{\sin }^2\theta \left[\text{using}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\right] \\ \text{=}{a}^2({\sin }^2\theta +{\cos }^2\theta )+b^2({\sin }^2\theta +{\cos }^2\theta ) \\ \text{=}{a}^2+b^2[\because {\sin }^2\theta +{\cos }^2\theta =1] \\ \text{=RHS} \\ \text{Hence proved.} \end{gathered}$$