Question

Simplify $2x^2+y^2+2xy=5$ where, $x=(2\cos \theta -\sin \theta )andy=(\cos \theta -3\sin \theta )$.

Answer 1

We have,

$2x^2+y^2+2xy=5$ $..........\left(1\right)$

Since,
$x=(2\cos \theta -\sin \theta )andy=(\cos \theta -3\sin \theta )$

So,

$2(2\cos \theta -\sin \theta {)}^2+(\cos \theta -3\sin \theta {)}^2+2\times (2\cos \theta -\sin \theta )\times (\cos \theta -3\sin \theta )=5$

$2(4{\cos }^2\theta +{\sin }^2\theta -4\cos \theta \sin \theta )+({\cos }^2\theta +9{\sin }^2\theta -6\cos \theta \sin \theta )+2(2{\cos }^2\theta -6\sin \theta \cos \theta -\sin \theta \cos \theta +3{\sin }^2\theta )=5$

$2(4{\cos }^2\theta +{\sin }^2\theta -4\cos \theta \sin \theta )+({\cos }^2\theta +9{\sin }^2\theta -6\cos \theta \sin \theta )+2(2{\cos }^2\theta -7\sin \theta \cos \theta +3{\sin }^2\theta )=5$

$8{\cos }^2\theta +2{\sin }^2\theta -8\cos \theta \sin \theta +{\cos }^2\theta +9{\sin }^2\theta -6\cos \theta \sin \theta +4{\cos }^2\theta -14\sin \theta \cos \theta +6{\sin }^2\theta =5$

$13{\cos }^2\theta +17{\sin }^2\theta -28\cos \theta \sin \theta =5$

$13({\cos }^2\theta +{\sin }^2\theta )+4{\sin }^2\theta -28\cos \theta \sin \theta =5$

$13+4{\sin }^2\theta -28\cos \theta \sin \theta =5$

$4{\sin }^2\theta -14\times 2\cos \theta \sin \theta =-8$

$4{\sin }^2\theta -14\sin 2\theta +8=0$

Hence, this is the answer.