Question

If $a\cos \theta -b\sin \theta =c$, show that
$\alpha \sin \theta +b\cos \theta =\pm \sqrt{a^2+b^2-c^2}$.

Answer 1

Square $a\cos \theta -b\sin \theta =c$. Then

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

or $a^2(1-{\sin }^2\theta )+b^2(1-{\cos }^2\theta )-2ab\sin \theta \cos \theta =c^2$

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

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

$\therefore a\sin \theta +b\cos \theta =\pm \sqrt{(a^2+b^2-c^2)}$