If sin is $\frac{\text{opposite}}{\text{hypotenuse}}$ then you know that the hypotenuse is a multiple of $2$ and the opposite side is half that. What kind of special right triangle fulfills those conditions?

$\text{tan}(x)=\frac{\text{sin}(x)}{\text{cos}(x)}$

$\text{sec}(x)=\frac{1}{\text{cos}(x)}$

$\text{csc}(x)=\frac{1}{\text{sin}(x)}$