Let θ = arctan ( 4 ) {\displaystyle \theta =\arctan(4)} . Then tan θ = 4 {\displaystyle \tan \theta =4} (since tan ( arctan ( θ ) ) = θ {\displaystyle \tan(\arctan(\theta ))=\theta } by definition of arctan {\displaystyle \arctan } ).
Draw a simple triangle expressing this — opposite=4, adjacent = 1, so hypotenuse = 4 2 + 1 2 = 17 {\displaystyle {\sqrt {4^{2}+1^{2}}}={\sqrt {17}}} ;
Then sin θ = opposite hypotenuse = 4 17 {\displaystyle \sin \theta ={\frac {\text{opposite}}{\text{hypotenuse}}}={\frac {4}{\sqrt {17}}}} .
To be sure of the sign, note that the arctangent of a positive number lies between 0 {\displaystyle 0} and π 2 {\displaystyle {\frac {\pi }{2}}} and the sine of such a number is positive.
Hence sin ( arctan ( 4 ) ) = 4 17 {\displaystyle \color {blue}\sin(\arctan(4))={\frac {4}{\sqrt {17}}}} .
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