Proof from Matrix

Question

I have tried upto where I could. please anyone help me to complete my proof..

Help much appreciated

It should only be a matter of plug in and calculate (as you have rightfully done), if I am not missing some preferable trick. I am sure there will be a matter of trig identities. — , Jun 17 '16 at 05:14
Do you know how to multiply the two matrices at the bottom of the page? — Greg Martin, Jun 17 '16 at 05:28
You could try using the identity $ \tan {\dfrac {\alpha }{2}}={\dfrac {1-\cos \alpha}{\sin \alpha}}$ — Joffan, Jun 17 '16 at 05:34

score 1 · Accepted Answer · answered Jun 17 '16 at 08:29

I therefore ends your calculations:

$\left( \begin{array}{cc} 1 & \frac{\sin \frac \alpha 2}{\cos \frac \alpha 2} \\ -\frac{\sin \frac \alpha 2}{\cos \frac \alpha 2} & 1 \end{array} \right) \left( \begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array} \right) =\left( \begin{array}{cc} \cos \alpha +\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\sin \alpha & -\sin \alpha +\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\cos \alpha \\ -\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\cos \alpha +\sin \alpha & \cos \alpha +\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\sin \alpha \end{array} \right) $ ;

so developing the expressions, in the last matrix, using trigonometric forms,

$\sin \alpha =2\sin \frac \alpha 2\cos \frac \alpha 2$

$\cos \alpha =2\cos ^2\frac \alpha 2-1$

$\cos ^2\frac \alpha 2+\sin ^2\frac \alpha 2=1$,

the result is obtained as $\left( \begin{array}{cc} 1 & -\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha } \\ \frac{\sin \frac 12\alpha }{\cos \frac 12\alpha } & 1 \end{array} \right) $

Proof from Matrix

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