[tex]\cos\dfrac{17\pi}8=\cos\dfrac\pi8[/tex]
Recall the half-angle identity:
[tex]\cos^2\dfrac\pi8=\dfrac{1+\cos\left(2\cdot\frac\pi8\right)}2\implies\cos\dfrac\pi8=\pm\sqrt{\dfrac{1+\cos\frac\pi4}2}[/tex]
[tex]\dfrac\pi8[/tex] lies in the first quadrant, which means [tex]\cos\dfrac\pi8>0[/tex] so we should take the positive square root. Then
[tex]\cos\dfrac\pi8=\sqrt{\dfrac{1+\frac1{\sqrt2}}2}=\dfrac{\sqrt{2+\sqrt2}}2[/tex]
