Rozwiązanie:
[tex]$L=\frac{1+\sin \alpha-\cos \alpha}{2\sqrt{2}\sin \Big(\frac{\alpha}{2}\Big)} =\frac{1+2\sin \Big(\frac{\alpha}{2}\Big)\cos \Big(\frac{\alpha}{2}\Big) -\Big(1-2\sin^{2} \Big(\frac{\alpha}{2}\Big)\Big)}{2\sqrt{2}\sin \Big(\frac{\alpha}{2}\Big)} =[/tex]
[tex]$=\frac{1+2\sin \Big(\frac{\alpha}{2}\Big) \cos \Big(\frac{\alpha}{2}\Big)+2\sin^{2} \Big(\frac{\alpha}{2}\Big)-1}{2\sqrt{2}\sin \Big(\frac{\alpha}{2}\Big)}=\frac{2\sin \Big(\frac{\alpha}{2}\Big)\cos \Big(\frac{\alpha}{2}\Big)+2\sin^{2} \Big(\frac{\alpha}{2}\Big)}{2\sqrt{2}\sin \Big(\frac{\alpha}{2}\Big)} =[/tex]
[tex]$=\frac{2\sin \Big(\frac{\alpha}{2}\Big)\Big(\cos \Big(\frac{\alpha}{2}\Big)+\sin \Big(\frac{\alpha}{2}\Big)\Big)}{2\sqrt{2}\sin \Big(\frac{\alpha}{2}\Big)} =\frac{\sin \Big(\frac{\alpha}{2}\Big)+\cos \Big(\frac{\alpha}{2}\Big)}{\sqrt{2}} =[/tex]
[tex]$=\frac{\sin \Big(\frac{\alpha}{2}\Big)+\sin \Big(\frac{\pi}{2}- \frac{\alpha}{2}\Big)}{\sqrt{2}} =\frac{2\sin \frac{\frac{\alpha}{2}+\frac{\pi}{2} -\frac{\alpha}{2} }{2} \cos \frac{\frac{\alpha}{2}-\frac{\pi}{2}+\frac{\alpha}{2} }{2} }{\sqrt{2}} =[/tex]
[tex]$=\frac{2 \sin \frac{\pi}{4} \cos \Big(\frac{\alpha}{2}-\frac{\pi}{4} \Big) }{\sqrt{2}} =\frac{2 \cdot \frac{\sqrt{2}}{2} \cos \Big(\frac{\alpha}{2}-\frac{\pi}{4} \Big) }{\sqrt{2}}=\frac{\sqrt{2} \cos \Big(\frac{\alpha}{2}-\frac{\pi}{4} \Big) }{\sqrt{2}}=[/tex]
[tex]$=\cos\Big(\frac{\alpha}{2}-\frac{\pi}{4} \Big)=\cos \Big(\frac{2\alpha-\pi}{4} \Big)=P[/tex]
co kończy dowód.
Założenia:
[tex]$\sin \frac{\alpha}{2} \neq 0 \iff \frac{\alpha}{2}\neq k\pi \iff \alpha \neq 2k\pi , \ \ k \in \mathbb{Z}[/tex]