Question

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

Answer 1

Given,

$a\cos \theta -b\sin \theta =c$

squaring on both sides, we get,

$(a\cos \theta -b\sin \theta )=c^2$

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

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

simplifying and rearranging the terms, we get,

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

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

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

Hence proved.