Bonjour Mwaka859
On utilise les formules :
[tex]\cos a=\sin(\dfrac{\pi}{2}-a)[/tex]
[tex]\sin(-a)=-\sin a[/tex].
[tex]\cos^2a+\sin^2a=1[/tex]
Nous avons ainsi :
[tex]\cos(x+\dfrac{\pi}{4})=sin(\dfrac{\pi}{2}-(x+\dfrac{\pi}{4}))\\\\\cos(x+\dfrac{\pi}{4})=sin(\dfrac{\pi}{2}-x-\dfrac{\pi}{4})\\\\\cos(x+\dfrac{\pi}{4})=sin(-x+\dfrac{\pi}{4})\\\\\cos(x+\dfrac{\pi}{4})=-sin(x-\dfrac{\pi}{4})\\\\cos^2(x+\dfrac{\pi}{4})=[-sin(x-\dfrac{\pi}{4})]^2\\\\\boxed{\cos^2(x+\dfrac{\pi}{4})=sin^2(x-\dfrac{\pi}{4})}[/tex]
[tex]\cos(x+\dfrac{5\pi}{4})=sin(\dfrac{\pi}{2}-(x+\dfrac{5\pi}{4}))\\\\\cos(x+\dfrac{5\pi}{4})=sin(\dfrac{\pi}{2}-x-\dfrac{5\pi}{4})\\\\\cos(x+\dfrac{5\pi}{4})=sin(-x-\dfrac{3\pi}{4})\\\\\cos(x+\dfrac{5\pi}{4})=-sin(x+\dfrac{3\pi}{4})\\\\cos^2(x+\dfrac{5\pi}{4})=[-sin(x+\dfrac{3\pi}{4})]^2\\\\\boxed{\cos^2(x+\dfrac{5\pi}{4})=sin^2(x+\dfrac{3\pi}{4})}[/tex]
D'où
[tex]\cos^2(x-\dfrac{\pi}{4})+\cos^2(x+\dfrac{\pi}{4})+\cos^2(x+\dfrac{3\pi}{4})+\cos^2(x+\dfrac{5\pi}{4})\\\\=\cos^2(x-\dfrac{\pi}{4})+\sin^2(x-\dfrac{\pi}{4})+\cos^2(x+\dfrac{3\pi}{4})+\sin^2(x+\dfrac{3\pi}{4})\\\\=[\cos^2(x-\dfrac{\pi}{4})+\sin^2(x-\dfrac{\pi}{4})]+[\cos^2(x+\dfrac{3\pi}{4})+\sin^2(x+\dfrac{3\pi}{4})]\\\\=1+1\\\\=2[/tex]
Par conséquent,
[tex]\\\\\boxed{\cos^2(x-\dfrac{\pi}{4})+\cos^2(x+\dfrac{\pi}{4})+\cos^2(x+\dfrac{3\pi}{4})+\cos^2(x+\dfrac{5\pi}{4})=2}[/tex]
