Question

If $\sin \theta =\frac{m}{\sqrt{m^2+n^2}}$ prove that $m\sin \theta +n\cos \theta =\sqrt{m^2+n^2}$

Answer 1

$\sin \theta =\frac{m}{\sqrt{m^2+n^2}}$

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

$\frac{m^2}{m^2+n^2}-1=-{\cos }^2\theta$

$-{\cos }^2\theta =\frac{m^2-m^2-n^2}{m^2+n^2}$

${\cos }^2\theta =\frac{n^2}{m^2+n^2}$

$\cos \theta =\frac{n}{\sqrt{m^2+n^2}}$

$m\sin \theta +n\cos \theta$

$m\left(\frac{m}{\sqrt{m^2+n^2}}\right)+n\left(\frac{n}{\sqrt{m^2+n^2}}\right)$

$\frac{m^2+n^2}{\sqrt{m^2+n^2}}=\sqrt{m^2+n^2}$

LHS=RHS