Questions mathématiques diverses

Question 53:
Montrer que si $\tan \dfrac{\theta}{2}=\sqrt{\dfrac{a-b}{a+b}}\tan \dfrac{\phi}{2}$
alors $\cos \theta=\dfrac{a\cos \phi+b}{a+b\cos \phi}$

Réponse 53:
$$\begin{align*} \cos \theta&=\dfrac{1-\tan^2\dfrac{\theta}{2}}{1+\tan^2\dfrac{\theta}{2}}\\ &=\dfrac{1-\dfrac{a-b}{a+b}\tan^2\dfrac{\phi}{2}}{1+\dfrac{a-b}{a+b}\tan^2\dfrac{\phi}{2}}\\ &=\dfrac{1-\dfrac{a-b}{a+b}\dfrac{\sin^2\frac{\phi}{2}}{\cos^2\frac{\phi}{2}}}{1+\dfrac{a-b}{a+b}\dfrac{\sin^2\frac{\phi}{2}}{\cos^2\frac{\phi}{2}}}\\ &=\dfrac{(a+b)\cos^2\dfrac{\phi}{2}-(a-b)\sin^2\dfrac{\phi}{2}}{(a+b)\cos^2\dfrac{\phi}{2}+(a-b)\sin^2\dfrac{\phi}{2}}\\ &=\dfrac{a\left({\cos^2\dfrac{\phi}{2}-\sin^2\dfrac{\phi}{2}}\right)+b\left({\cos^2\dfrac{\phi}{2}+\sin^2\dfrac{\phi}{2}}\right)}{a\left({\cos^2\dfrac{\phi}{2}+\sin^2\dfrac{\phi}{2}}\right)+b\left({\cos^2\dfrac{\phi}{2}-\sin^2\dfrac{\phi}{2}}\right)}\\ &=\dfrac{a\cos\phi+b}{a+b\cos\phi} \end{align*}$$