Как найти sin 13пи/8

[latex]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}[/latex] Ответ: [latex]-\dfrac{\sqrt{2+\sqrt2}}{2}[/latex]

 sin(13π/8)=sin(π+5π/8)=- sin5π/8=- sin(π/2+π/8)=- cosπ/8=- cos(22,5град)= sin(13π/8)= - 0,9239