Answer:
[tex]\frac{1}{2}\sqrt{2+\sqrt{3}}[/tex]
Step-by-step explanation:
we know that
An half-angle identity is equal to
[tex]sin(\frac{\theta}{2})=(+/-)\sqrt{\frac{1-cos(\theta)}{2}}[/tex]
we have
[tex]sin(\frac{5\pi}{12})[/tex]
The angle [tex]\frac{5\pi}{12}=75^o[/tex] ----> belong to the First Quadrant, so the value of the sine is positive
Let
[tex]\frac{\theta}{2}=\frac{5\pi}{12}[/tex]
so
[tex]{\theta=\frac{5\pi}{6}[/tex]
[tex]sin(\frac{5\pi}{12})=\sqrt{\frac{1-cos(\theta)}{2}}[/tex]
[tex]cos(\theta)=cos(\frac{5\pi}{6})=cos(150^o)=-\frac{\sqrt{3}}{2}[/tex]
substitute
[tex]sin(\frac{5\pi}{12})=\sqrt{\frac{1-(-\frac{\sqrt{3}}{2})}{2}}[/tex]
[tex]sin(\frac{5\pi}{12})=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}[/tex]
[tex]sin(\frac{5\pi}{12})=\sqrt{\frac{2+\sqrt{3}}{4}[/tex]
[tex]sin(\frac{5\pi}{12})=\frac{1}{2}\sqrt{2+\sqrt{3}}[/tex]