7π/12 lies in the second quadrant, so we expect cos(7π/12) to be negative.
Recall that
[tex]\cos^2x=\dfrac{1+\cos(2x)}2[/tex]
which tells us
[tex]\cos\left(\dfrac{7\pi}{12}\right)=-\sqrt{\dfrac{1+\cos\left(\frac{7\pi}6\right)}2}[/tex]
Now,
[tex]\cos\left(\dfrac{7\pi}6\right)=-\cos\left(\dfrac\pi6\right)=-\dfrac{\sqrt3}2[/tex]
and so
[tex]\cos\left(\dfrac{7\pi}{12}\right)=-\sqrt{\dfrac{1-\frac{\sqrt3}2}2}=\boxed{-\dfrac{\sqrt{2-\sqrt3}}2}[/tex]
