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coolkid4
wt
it is easiest to do it with a picture of a triangle |dw:1354676626218:dw|
that is a picture of a triangle with angle \(x\) and with \(\tan(x)=2\) so \(x=\arctan(2)\) and you are looking for \(\sin(\arctan(2))=\sin(\arctan(x))\) to find it you need the hypotenuse of the triangle. by pythagoras it is \[\sqrt{2^2+1^2}=\sqrt{5}\]
sorry i meant you are looking for \(\sin(x)\)
now we have completed the triangle |dw:1354676824808:dw|
and \(\sin(x)=\sin(\arctan(2))=\frac{2}{\sqrt{5}}\)
yw, sorry about the typo