Exercise 6.5.26

Here we have

and

1.

Compute

$${[\mathit{UT}]}_{\alpha }={[U]}_{\alpha }{[T]}_{\alpha }=\left(\begin{array}{cc}\hfill \cos <!--\; FUNCTION\; APPLICATION\; -->2(\psi -\frac{\varphi }{2})\hfill & \hfill \sin <!--\; FUNCTION\; APPLICATION\; -->2(\psi -\frac{\varphi }{2})\hfill \\ \hfill \sin <!--\; FUNCTION\; APPLICATION\; -->2(\psi -\frac{\varphi }{2})\hfill & \hfill -\cos <!--\; FUNCTION\; APPLICATION\; -->2(\psi -\frac{\varphi }{2})\hfill \end{array}\right).$$

So the angle is $\psi -\frac{\varphi }{2}$.

2.

Compute

$${[\mathit{TU}]}_{\alpha }={[T]}_{\alpha }{[U]}_{\alpha }=\left(\begin{array}{cc}\hfill \cos <!--\; FUNCTION\; APPLICATION\; -->2(\psi +\frac{\varphi }{2})\hfill & \hfill \sin <!--\; FUNCTION\; APPLICATION\; -->2(\psi +\frac{\varphi }{2})\hfill \\ \hfill \sin <!--\; FUNCTION\; APPLICATION\; -->2(\psi +\frac{\varphi }{2})\hfill & \hfill -\cos <!--\; FUNCTION\; APPLICATION\; -->2(\psi +\frac{\varphi }{2})\hfill \end{array}\right).$$

So the angle is $\psi +\frac{\varphi }{2}$.