\(\cos\left(-\dfrac{7\pi}{12}\right)=\cos\dfrac{7\pi}{12}\)
Recall that
\(\cos^2x=\dfrac{1+\cos2x}2\)
Let \(x=\dfrac{7\pi}{12}\), so that
\(\cos^2\dfrac{7\pi}{12}=\dfrac{1+\cos\dfrac{7\pi}6}2=\dfrac{1-\dfrac{\sqrt3}2}2=\dfrac{2-\sqrt3}4\)
\(\cos\dfrac{7\pi}{12}=\pm\sqrt{\dfrac{2-\sqrt3}4}=\pm\dfrac{\sqrt{2-\sqrt3}}2\)
Since the angle \(\dfrac{7\pi}{12}\) lies in the second quadrant, you know that the cosine must be negative, so
\(\cos\dfrac{7\pi}{12}=-\dfrac{\sqrt{2-\sqrt3}}2\)
