Question

Find the exact value by using a half-angle identity. sin(7" title=" \pi " alt=" \pi " align="absmiddle" class="latex-formula">/8)

Answer 1

$\bf sin\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1-cos({{ \theta}})}{2}}\\\\ -------------------------------\\\\ \cfrac{7\pi }{8}\cdot 2\implies \cfrac{7\pi }{4}\qquad \qquad sin\left(\frac{7\pi }{8} \right)\implies sin\left(\cfrac{\frac{7\pi }{4}}{2} \right) \\\\\\ sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{1-cos\left(\frac{7\pi }{4} \right)}{2}}\implies sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{1-\frac{\sqrt{2}}{2}}{2}}$

$\bf sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{\frac{2-\sqrt{2}}{2}}{2}} \implies sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{2-\sqrt{2}}{4}} \\\\\\ sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies sin\left(\cfrac{\frac{7\pi }{4}}{2} \right)=\pm\cfrac{\sqrt{2-\sqrt{2}}}{2}$