Question

Prove that (cosA+cosB/sinA-sinB)^n + (sinA+sinB/cosA-cosB)^n ={2cot^n(A-B/2),if even}

{0, if odd}

Answer 1

$$\cos a + \cos b = 2 \cos \frac {a + b}{2} \cos \frac {a - b}{2}$$ $$\sin a \pm \sin b = 2 \sin \frac {a \pm b}{2} \cos \frac {a \mp b}{2}$$ $$\cos a - \cos b = - 2 \sin \frac {a + b}{2} \sin \frac {a - b}{2}$$ $$\left(\frac {\cos a + \cos b}{\sin a - \sin b}\right) ^ {n} + \left(\frac {\sin a + \sin b}{\cos a - \cos b}\right) ^ {n} = \left(\frac {2 \cos \frac {a + b}{2} \cos \frac {a - b}{2}}{2 \sin \frac {a - b}{2} \cos \frac {a + b}{2}}\right) ^ {n} + \left(\frac {2 \sin \frac {a + b}{2} \cos \frac {a - b}{2}}{- 2 \sin \frac {a + b}{2} \sin \frac {a - b}{2}}\right) ^ {n} =$$ $$= \left(1 + (- 1) ^ {n}\right) \cot^ {n} \left(\frac {a - b}{2}\right) = \left\{ \begin{array}{l} 2 \cot^ {n} \frac {a - b}{2}, n = 2 k \\ 0, n = 2 k + 1 \end{array} \right.$$