Bonsoir,
On sait que, quel que soit x, on a :
[tex]\cos ^2x + \sin ^2x = 1[/tex]
Donc, on a :
[tex]\sin^2 x = 1-\cos ^2x\\ \sin x = \sqrt{1-\cos^2x}\\ \sin x = \sqrt{1-\left(\frac 12\right) ^2}\\ \sin x = \sqrt{1-\frac 14} = \sqrt{\frac 34} = \frac{\sqrt 3}{2}[/tex]
