If tan(x/2) = m/n, Then Find m sin x + n cos x
Question:
\[ \tan\frac{x}{2}=\frac{m}{n} \] Find \[ m\sin x+n\cos x. \]Solution
Using the identities
\[ \sin x=\frac{2\tan(x/2)}{1+\tan^2(x/2)} \] and \[ \cos x=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)} \]Since
\[ \tan\frac{x}{2}=\frac{m}{n}, \]we get
\[ \sin x = \frac{2(m/n)}{1+m^2/n^2} = \frac{2mn}{m^2+n^2} \] and \[ \cos x = \frac{1-m^2/n^2}{1+m^2/n^2} = \frac{n^2-m^2}{m^2+n^2}. \]Therefore,
\[ m\sin x+n\cos x = m\left(\frac{2mn}{m^2+n^2}\right) + n\left(\frac{n^2-m^2}{m^2+n^2}\right) \] \[ = \frac{2m^2n+n^3-m^2n}{m^2+n^2} \] \[ = \frac{m^2n+n^3}{m^2+n^2} \] \[ = \frac{n(m^2+n^2)}{m^2+n^2} \] \[ =n. \]