[tex]\bf 2\cdot \cfrac{3}{8}\implies \cfrac{3}{4}\qquad thus
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sin\left( \cfrac{3\pi }{8} \right)\iff sin\left( \cfrac{\frac{3\pi }{4}}{2} \right)=\pm \sqrt{\cfrac{1-cos\left( \frac{3\pi }{4} \right)}{2}}
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sin\left( \cfrac{\frac{3\pi }{4}}{2} \right)=\pm \sqrt{\cfrac{1-\left( -\frac{\sqrt{2}}{2} \right)}{2}}\implies sin\left( \cfrac{\frac{3\pi }{4}}{2} \right)=\pm \sqrt{\cfrac{\frac{2+\sqrt{2}}{2}}{2}}
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sin\left( \cfrac{\frac{3\pi }{4}}{2} \right)=\pm\sqrt{\cfrac{2+\sqrt{2}}{4}}[/tex]
