Respuesta :

lastbenchstudent
let assume

[tex]sin( {tan}^{ - 1} 5) = sin \: y \\ so \: y \: = {tan}^{ - 1} 5 \: \\ tan \: y \: = 5 \\ [/tex]

we know from defination of inverse function

tan y = opposite / adjacent

here opposite is 5 and adjacent is 1

tan y = 5/1

lets calculate hypotenus using pythagoras theorem

hypotenuse = root ( 5^2 + 1^2)= root (25+1) = root (26)

now we know

sin y = opposite / hypotenus

so

[tex]sin \: y \: = \frac{5}{ \sqrt{26} } \\ sin \: ( {tan}^{ - 1} x)\: = \frac{5}{ \sqrt{26} } [/tex]

hope it helps you

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