Answer :
[tex]\bf sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-cos({{ \theta}})}{2}}\\\\
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\cfrac{7\pi }{8}\cdot 2\implies \cfrac{7\pi }{4}\qquad \qquad sin\left(\frac{7\pi }{8} \right)\implies sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)
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sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{1-cos\left(\frac{7\pi }{4} \right)}{2}}\implies
sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{1-\frac{\sqrt{2}}{2}}{2}}[/tex]
[tex]\bf sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{\frac{2-\sqrt{2}}{2}}{2}} \implies sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{2-\sqrt{2}}{4}} \\\\\\ sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\cfrac{\sqrt{2-\sqrt{2}}}{2}[/tex]
[tex]\bf sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{\frac{2-\sqrt{2}}{2}}{2}} \implies sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{2-\sqrt{2}}{4}} \\\\\\ sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\cfrac{\sqrt{2-\sqrt{2}}}{2}[/tex]