Chapter 5 - Section 5.4 - Half-Angle Formulas - 5.4 Problem Set - Page 304: 7

False

Testing the equation: $\sin{\dfrac{A}{2}} = \dfrac{1}{2} \sin{A}$ for $A=60^\circ$ $LHS = \sin{30} = \dfrac{1}{2}$ $RHS = \dfrac{1}{2}\sin{60} = \dfrac{\sqrt{3}}{4}$ $LHS \neq RHS$ $\therefore \sin{\dfrac{A}{2}} \neq \dfrac{1}{2}\sin{A}$