Answer:
Step-by-step explanation:
Remember:
[tex]cos(2A)=cos(A+A)=cos^2(A)-sin^2(A)=2cos^2(A)-1\\\\cos^2(A)=\dfrac{cos(2A)+1}{2} \\\\cos(A)=\pm\sqrt{\dfrac{cos(2A)+1}{2}} \\\\[/tex]
[tex]cos (\dfrac{A}{2} )=\pm\sqrt{\dfrac{1+cos(A)}{2} } \ with \left\{\begin{array}{c} \ +:\ if \frac{A}{2} \ is\ in\ I\ or\ IV\ quadran\\\ -:\ if \frac{A}{2} \ is\ in\ II\ or\ III\ quadran\\\end{array}\right.[/tex]
[tex]cos(\pi)=-1\\\\cos(\dfrac{\pi}{2} )=+\sqrt{\dfrac{1+(-1)}{2}} =0\\\\cos(\dfrac{\pi}{4} )=+\sqrt{\dfrac{1+0)}{2}} =\dfrac{\sqrt{2} }{2} \\[/tex]
[tex]cos(\dfrac{\pi}{8} )=\sqrt{\dfrac{1+\dfrac{\sqrt{2}}{2})}{2}} =\sqrt{\dfrac{1}{2} +\dfrac{\sqrt{2} }{4} }=\dfrac{1}{2} \sqrt{2+\sqrt{2} } \\\\\\cos(\dfrac{\pi}{16} )=\sqrt{\dfrac{1+\dfrac{1}{2}*\sqrt{2+\sqrt{2} } }{2}}\\\\\\=\sqrt{\frac{1}{2} +\frac{1}{4}*\sqrt{2+\sqrt{2} } } \\\\\\=\dfrac{1}{2}*\sqrt{2+\sqrt{2+\sqrt{2} } } \\\\[/tex]
