Question

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

Answer 1

We have, $a\cos \theta -b\sin \theta =c.$

On squaring both sides, we get

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

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

[Using the identity, $(a-b{)}^2=a^2+b^2-2ab$]

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

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

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

[Using the identity, ${\sin }^2\theta +{\cos }^2\theta =1,{\sin }^2\theta =1-{\cos }^2\theta$ and ${\cos }^2\theta =1-{\sin }^2\theta$]

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

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

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

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

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

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

[Using the identity, $a^2+b^2+2ab=(a+b{)}^2$ ]

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