zagonek34

2021-12-31

How to solve $\frac{1}{2{\mathrm{sin}50}^{\circ }}+2{\mathrm{sin}10}^{\circ }$

Paul Mitchell

$\frac{1}{2{\mathrm{sin}50}^{\circ }}+2{\mathrm{sin}10}^{\circ }=\frac{1+4{\mathrm{sin}50}^{\circ }{\mathrm{sin}10}^{\circ }}{2{\mathrm{sin}50}^{\circ }}$
$=\frac{1+2\left({\mathrm{cos}40}^{\circ }-{\mathrm{cos}60}^{\circ }\right)}{2{\mathrm{cos}40}^{\circ }}$
=1

David Clayton

$\frac{1}{2{\mathrm{sin}50}^{\circ }}+2{\mathrm{sin}10}^{\circ }=\frac{1}{2{\mathrm{sin}50}^{\circ }}+2\mathrm{sin}\left({60}^{\circ }-{50}^{\circ }\right)=\frac{1}{2{\mathrm{sin}50}^{\circ }}+\sqrt{3}{\mathrm{cos}50}^{\circ }-{\mathrm{sin}50}^{\circ }=$
$=\frac{1-2{\mathrm{sin}}^{2}{50}^{\circ }}{2{\mathrm{sin}50}^{\circ }}+\frac{2\sqrt{3}{\mathrm{cos}50}^{\circ }{\mathrm{sin}50}^{\circ }}{2{\mathrm{sin}50}^{\circ }}=\frac{{\mathrm{cos}100}^{\circ }}{2{\mathrm{sin}50}^{\circ }}+\frac{\sqrt{3}{\mathrm{sin}100}^{\circ }}{2{\mathrm{sin}50}^{\circ }}=$
$=\frac{{\frac{12}{\mathrm{cos}100}}^{\circ }+\frac{\sqrt{3}}{2}{\mathrm{sin}100}^{\circ }}{{\mathrm{sin}50}^{\circ }}=\frac{\mathrm{sin}\left({30}^{\circ }+{100}^{\circ }\right)}{{\mathrm{sin}50}^{\circ }}=1$

karton

Always remember : "never simplify denominator unless you know the values."
just cross multiply to get
$\frac{1+4\mathrm{sin}50\mathrm{sin}10}{2\mathrm{sin}50}$
which would further simplify as
$\frac{1+2\left(\mathrm{cos}40-\mathrm{cos}60\right)}{2\mathrm{sin}50}$
which is easy to solve as we get
$\frac{\mathrm{cos}40}{\mathrm{sin}50}=1$