+9665 \(\tan t=\frac{12}{5}\\ hypotenuse = \sqrt{5^2+12^2}=13\\ \cos \space t= \frac{5}{13} \\\quad\sin(\frac{t}{2})\\= \pm\sqrt{\frac{1-\cos t}{2}}\\=\pm\sqrt{\frac{1-\frac{5}{13}}{2}}\\=\pm\sqrt{\frac{\frac{8}{2}}{13}}\\\)
\(=\pm\sqrt{\frac{4}{13}}\\=\frac{\pm2}{\sqrt{13}}\)
\(\because \space \tan t \space is \space positive\\\therefore \sin(\frac{t}{2})\space is\space positive\space too\)
\(\therefore \sin(\frac{t}{2})=\frac{2\sqrt{13}}{13}\space OR \frac{2}{\sqrt{13}}\)
By the way, why pie?
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