Answer:
[tex]\frac{\sqrt 6 +\sqrt 2}{4}[/tex]
Step-by-step explanation:
Use the addition formula [tex]\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)[/tex] with x=4π/12 and y=3π/12:
[tex] \sin(\frac{7\pi}{12})&=\sin(\frac{4\pi}{12}+\frac{3\pi}{12})=\sin(\frac{4\pi}{12})\cos(\frac{3\pi}{12})+\cos(\frac{4\pi}{12})\sin(\frac{3\pi}{12})=\sin(\frac{pi}{3})\cos(\frac{\pi}{4})+\cos(\frac{\pi}{3})\sin(\frac{\pi}{4})=\frac{\sqrt 3}{2}\cdot\frac{\sqrt 2}{2}+\frac{1}{2}\cdot\frac{\sqrt 2}{2}=\frac{\sqrt 6 +\sqrt 2}{4}[/tex]
