KTOM Indonesia, October 2020
$\sin(5^{\circ})\sin(55^{\circ})\sin(65^{\circ})=\displaystyle\frac{A\sqrt{B}+C\sqrt{D}}{E}$
where $ B $ and $ D $ are two natural numbers that are not divisible by the square of any prime, $ A $ and $ C $ are integers, and $ E $ is a natural number. Find the value of $ A + B + C + D + E. $
Continuing from what others said.... $$sin(5^{\circ})sin(55^{\circ})sin(65^{\circ})=\frac{sin(15^{\circ})}{4}$$
$$cos(30^{\circ})=1-2sin^{2}(15^{\circ})$$ $$2sin^{2}(15^{\circ})=1-cos(30^{\circ})$$ $$2sin^{2}(15^{\circ})=1-cos(30^{\circ})$$ $$2sin^{2}(15^{\circ})=1-\frac{\sqrt3}{2}$$ $$2sin^{2}(15^{\circ})=\frac{2-\sqrt3}{2}$$ $$sin^{2}(15^{\circ})=\frac{2-\sqrt3}{4}$$ $$sin(15^{\circ})=\frac{\sqrt{2-\sqrt3}}{2}=\frac{\sqrt{\biggl(\sqrt\frac{1}{2}-\sqrt\frac{3}{2}\biggr)^{2}}}{2}$$ $$=\frac{\biggr(\sqrt\frac{3}{2}-\sqrt\frac{1}{2}\biggl)}{2}\ \ (as\ sin(15^{\circ})\ is\ always\ positive)$$ $$\Rightarrow \frac{sin(15^{\circ})}{4}=\frac{\biggr(\sqrt\frac{3}{2}-\sqrt\frac{1}{2}\biggl)}{8}=\biggr(\frac{\sqrt3-1}{8\sqrt2}\biggl)=\biggr(\frac{\sqrt6-\sqrt2}{16}\biggl)$$ $$\Rightarrow\frac{A\sqrt{B}+C\sqrt{D}}{E}=\biggr(\frac{1\sqrt6-1\sqrt2}{16}\biggl)$$ 1 and-1 are integers, 2 and 6 are natural numbers which is not divisible by square of any prime number and 16 is a natural number.
So, $$A+B+C+D+E=1+6-1+2+16=24$$
Use this identity- $$sin(\theta)\cdot sin(60-\theta)\cdot sin(60+\theta)=\frac{1}{4}sin(3\theta)$$
