Question

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

Answer 1

given $a\cos \left(\theta \right)+b\sin \left(\theta \right)=c$

let $a\sin \left(\theta \right)-b\cos \left(\theta \right)=x$

now squaring and adding both equations we get

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

therefore

$x=\pm \sqrt{a^2+b^2-c^2}$

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