If $a\cos\theta+b\sin \theta=m$ and $a\sin \theta-b\cos\theta=n$ then show that $a^2+b^2=m^2+n^2$

Answer 1

Toolbox:

• $\sin^2\theta+\cos^2\theta=1$

$a\cos\theta+b\sin \theta=m$

$(a\cos\theta+b\sin \theta)^2=m^2$

$a^2\cos^2\theta+b^2\sin^2\theta+2ab\sin \theta\cos\theta=m^2$-----(1)

$a\sin\theta+b\cos \theta=n$

$(a\sin\theta+b\cos \theta)^2=n^2$

$a^2\sin^2\theta+b^2\cos^2\theta-2ab\sin \theta\cos\theta=n^2$-----(2)

After adding (1) & (2)

$a^2\cos^2\theta+b^2\sin^2\theta+a^2\sin^2\theta+b^2\cos^2\theta=m^2+n^2$

$a^2(\cos^2\theta+\sin^2\theta)+b^2(\cos^2\theta+\sin^2\theta)=m^2+n^2$

$\sin^2\theta+\cos^2\theta=1$

$a^2+b^2=m^2+n^2$

Hence proved.