Exercise 5.4.5 solution
Since \(A\) is in quadrant IV, \(x\) must be positive and and \(y\) must be negative.
\(\tan\frac{A}{2}=\frac{\sin A}{1+\cos A}=\dfrac{-\frac{3}{5}}{1+\frac{4}{5}}=-\frac{1}{3}\)
Since \(A\) is in quadrant IV, \(x\) must be positive and and \(y\) must be negative.
\(\tan\frac{A}{2}=\frac{\sin A}{1+\cos A}=\dfrac{-\frac{3}{5}}{1+\frac{4}{5}}=-\frac{1}{3}\)