Question

If $\cos \theta =\frac{2\sqrt{mn}}{m+n}$, find the value of $\sin \theta$ (given $m>n$)

Answer 1

Given

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

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

$\Rightarrow {\sin }^2\theta =1-\frac{4mn}{(m+n{)}^2}$

$\Rightarrow {\sin }^2\theta =\frac{m^2+n^2+2mn-4mn}{(m+n{)}^2}$

$\Rightarrow {\sin }^2\theta =\frac{(m-n{)}^2}{(m+n{)}^2}$

$\therefore \sin \theta =\frac{m-n}{m+n}$