Bonjour Ylkcomtois27
[tex]1)\ A(x)=\cos(-x)-\sin(\pi+x)+\sin(-x)+\cos(\pi-x)\\\\\cos(-x)=\cos(x)\\\sin(\pi+x)=-\sin(x)\\\sin(-x)=-\sin(x)\\\cos(\pi-x)=-\cos(x)\\\\\Longrightarrow A(x)=\cos(x)+\sin(x)-\sin(x)-\cos(x)\\\\\Longrightarrow\boxed{A(x)=0}\\\\B(x)=\sin(-x)-\cos(\pi+x)-\cos(-x)+\sin(\pi-x)\\\\\sin(-x)=-\sin(x)\\\cos(\pi+x)=-\cos(x)\\\cos(-x)=\cos(x)\\\sin(\pi-x)=\sin(x)\\\\\Longrightarrow B(x)=-\sin(x)+\cos(x)-\cos(x)+\sin(x)\\\\\Longrightarrow\boxed{B(x)=0}[/tex]
[tex]2) \sin^2(\dfrac{\pi}{5})+\cos^2(\dfrac{\pi}{5})=1\\\\\Longrightarrow\sin^2(\dfrac{\pi}{5})=1-\cos^2(\dfrac{\pi}{5})\\\\\Longrightarrow\sin^2(\dfrac{\pi}{5})=1-(\dfrac{1+\sqrt{5}}{4})^2\\\\\Longrightarrow\sin^2(\dfrac{\pi}{5})=1-\dfrac{1+2\sqrt{5}+(\sqrt{5})^2}{16}\\\\\Longrightarrow\sin^2(\dfrac{\pi}{5})=1-\dfrac{1+2\sqrt{5}+5}{16}\\\\\Longrightarrow\sin^2(\dfrac{\pi}{5})=1-\dfrac{6+2\sqrt{5}}{16}\\\\\Longrightarrow\sin^2(\dfrac{\pi}{5})=\dfrac{16-6-2\sqrt{5}}{16}[/tex]
[tex]\\\\\Longrightarrow\sin^2(\dfrac{\pi}{5})=\dfrac{10-2\sqrt{5}}{16}\\\\\Longrightarrow\sin(\dfrac{\pi}{5})=\sqrt{\dfrac{10-2\sqrt{5}}{16}}\ \ \ (car\ \ \sin(\dfrac{\pi}{5})\ \textgreater \ 0)\\\\\\\Longrightarrow\boxed{\sin(\dfrac{\pi}{5})=\dfrac{\sqrt{10-2\sqrt{5}}}{4}}[/tex]
[tex]\boxed{\sin(\dfrac{6\pi}{5})=-\sin(\dfrac{\pi}{5})=\dfrac{-\sqrt{10-2\sqrt{5}}}{4}}\\\\\boxed{\cos(\dfrac{4\pi}{5})=-\cos(\dfrac{\pi}{5})=\dfrac{-1-\sqrt{5}}{4}}[/tex]
