Question

Să se demonstreze că:
1) sin $\frac{a}{2}$=+ sau - $\sqrt{ \frac{1-cos a}{2} }$
2) cos $\frac{a}{2}$=+ sau - $\sqrt{ \frac{1+cos a }{2} }$

Answer 1

[tex]1) \cos2a=\cos a\cdot \cos a-\sin a\cdot \sin a\\ \cos 2a=\cos^2a-\sin^2a\\ \cos 2a=1-2\sin ^2a\\ 2\sin^2a =1-\cos 2a\\ \sin^2a=\dfrac{1-\cos 2a}{2}\\ |\sin a|=\sqrt{\dfrac{1-\cos 2a}{2}}\\ \left|\sin \dfrac{a}{2}\right|=\sqrt{\dfrac{1-\cos a}{2}}\\ \text{Asadar:}\\ \sin \dfrac{a}{2}=\pm \sqrt{\dfrac{1-\cos a}{2}} [/tex]
[tex]2)\text{Se face la fel ca si la 1).}\\ \cos 2x=\cos^2x-\sin^2x\\ \cos 2x=2\cos^2x-1\\ 2\cos^2x=\cos2x+1\\ cos^2x=\dfrac{\cos2x+1}{2}\\ | \cos x|=\sqrt{\dfrac{\cos 2x+1}{2}}\\ \left |\cos \dfrac{x}{2}\right |=\sqrt{\dfrac{\cos x+1}{2}}\\ Deci:\\ \cos \dfrac{x}{2}=\pm \sqrt{\dfrac{\cos x+1}{2}}[/tex]