## Precalculus (6th Edition)

$\displaystyle \sin\frac{3\theta}{10}$
Comparing with the Half-Angle Identity, $\displaystyle \sin\frac{A}{2}=\pm\sqrt{\frac{1-\cos A}{2}}$ if we replace $A$ with $\displaystyle \frac{3\theta}{5}$ in the identity, the RHS equals the given expression. So, the LHS $=\displaystyle \sin\frac{A}{2}$= $\displaystyle \sin\frac{\frac{3\theta}{5}}{2}=\sin\frac{3\theta}{10}$