[tex]sin \frac{13 \pi }{8} =sin( \pi + \frac{5 \pi }{8})=-sin \frac{5 \pi }{8} = -sin( \frac{\pi}{2} + \frac{ \pi }{8})=-cos \frac{ \pi }{8}\\ \\
cos \frac{ \pi }{8}=\sqrt{\dfrac{1+cos \frac{ \pi }{4} }{2}} = \sqrt{\dfrac{1+\frac{ \sqrt2}{2} }{2}}=\dfrac{\sqrt{2+\sqrt2}}{2} \\ \\
sin \frac{13 \pi }{8} =-\dfrac{\sqrt{2+\sqrt2}}{2}[/tex]
Ответ: [tex]-\dfrac{\sqrt{2+\sqrt2}}{2}[/tex]
