Bonjour EMELINE2108
[tex]1)\ \sin x=-\dfrac{1}{3}\ (x\in[\dfrac{\pi}{2};\dfrac{3\pi}{2}]\\\\\cos^2x+\sin^2x=1\\\\\cos^2x=1-\sin^2x\\\\\cos^2x=1-(-\dfrac{1}{3})^2\\\\\cos^2x=1-\dfrac{1}{9}\\\\\cos^2x=\dfrac{8}{9}\\\\\cos x=\pm\sqrt{\dfrac{8}{9}}=\pm\dfrac{2\sqrt{2}}{3}}\\\\Or\ x\in[\dfrac{\pi}{2};\dfrac{3\pi}{2}]\Longrightarrow\cos x\le0\\\\Donc\ \boxed{\cos x=-\dfrac{2\sqrt{2}}{3}}}[/tex]
[tex]2)\ \sin x =-\dfrac{1}{3}\Longrightarrow x\approx-0,34+2k\pi\ \ ou\ \ x\approx\pi-(-0,34)+2k\pi\\\\\sin x =-\dfrac{1}{3}\Longrightarrow x\approx-0,34+2k\pi\ \ ou\ \ x\approx\pi+0,34+2k\pi\\\\Or\ x\in[\dfrac{\pi}{2};\dfrac{3\pi}{2}]\Longrightarrow x\approx\pi+0,34\approx3,48\\\\\boxed{x\approx3,48}[/tex]
3) cos(x + 7pi/2) = cos(x + 7pi/2 - 8pi/2) = cos(x - pi/2) = cos(pi/2 - x) = sin x
sin (x - 5pi) = sin(x - pi - 4pi) = sin(x - pi) = - sin(pi - x) = -sin x
cos (2016pi - x) = cos(-x) = cos x
sin (5pi/2 -x) = sin(4pi/2 + pi/2 - x) = sin(pi/2 - x) = cos x
