Question

If $x=asin\theta +bcos\theta$ and $y=acos\theta -bsin\theta ,$ prove that $x^2+y^2=a^2+b^2$

Answer 1

Step 1:- Given that,

$x=asin\theta +bcos\theta$

and $y=acos\theta -bsin\theta$

Step 2: We have to prove that ${\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{y}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{b}}^{\mathbf{2}}$

Now L.H.S.$=x^2+y^2$

$={\left(a\sin \theta +b\cos \theta \right)}^2+{\left(a\cos \theta -b\sin \theta \right)}^2$

$=a^2{\sin }^2\theta +b^2{\cos }^2\theta +2ab\sin \theta \cos \theta +a^2{\cos }^2\theta +b^2{\sin }^2\theta -2ab\sin \theta \cos \theta$

$=a^2\left({\sin }^2\theta +{\cos }^2\theta \right)-b^2\left({\sin }^2\theta -{\cos }^2\theta \right)$ $\therefore {\sin }^2\theta -{\cos }^2\theta$

$=a^2+b^2$

$=p^2-q^2=\mathrm{R}.\mathrm{H}.\mathrm{S}$

Hence, L.H.S. = R.H.S Proved.