Solution
= Solution
Put $\theta=\pi/12$. The double-angle <trigonometric identities> imply $\cos^2(2\theta)=1-\sin^2(2\theta)$. Since $2\theta=\pi/6$ is in the first quadrant, the positive <square root> is required. The <half-angle identities> then give the requested expression in terms of $\sin(\pi/6)$:
$$
\boxed{\sin^2\frac\pi{12}=\frac{1-\sqrt{1-\sin^2(\pi/6)}}2.}
$$