Mathematics, 03.02.2020 23:54 daelinrobinson
Use a half-angle identity to find the exact value of sin75
Answers: 3
![\bf \textit{Half-Angle Identities} \\\\ sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 75\cdot 2\implies 150\qquad \qquad \qquad \cfrac{150}{2}\implies 75 \\\\[-0.35em] ~\dotfill\\\\ sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{1-cos(150^o)}{2}}\implies sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{1-\left( -\frac{\sqrt{3}}{2} \right)}{2}}](/tpl/images/0498/3036/e7be0.png)
![\bf sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}} \\\\\\ sin\left( \cfrac{150^o}{2} \right)=\pm\sqrt{\cfrac{2+\sqrt{3}}{4}}\implies sin\left( \cfrac{150^o}{2} \right)=\pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill sin\left( \cfrac{150^o}{2} \right)=\pm\cfrac{\sqrt{2+\sqrt{3}}}{2}~\hfill](/tpl/images/0498/3036/da6c0.png)
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