Respuesta :
Answer:
[tex]\frac{x}{\sqrt{x^2+9}}[/tex]
Step-by-step explanation:
Let [tex]u=\arctan(\frac{x}{3})[/tex].
Then [tex]\tan(u)=\frac{x}{3}[/tex].
Construct a right triangle.
Make one of the non-right angles [tex]u[/tex].
Recall [tex]\tan(u)[/tex] is equal to the ratio of the side that is opposite to [tex]u[/tex] to the side that is adjacent to [tex]u[/tex].
This means opposite side [tex]=x[/tex] while adjacent side[tex]=3[/tex] since [tex]\tan(u)=\frac{x}{3}=\frac{\text{opposite}}{\text{adjacent}}[/tex].
We can find the hypotenuse by using the Pythagorean Theorem.
[tex](x)^2+(3)^2=(\text{hyp})^2[/tex]
[tex]x^2+9=(\text{hyp})^2[/tex]
Take the square root of both sides:
[tex]\sqrt{x^2+9}=\text{hyp}[/tex].
Recall [tex]\sin(u)[/tex] is equal to the ratio of the side that is opposite to [tex]u[/tex] to the hypotenuse.
[tex]\sin(\arctan(\frac{x}{3}))[/tex]
[tex]\sin(u)[/tex] -->Because we let the inside equal [tex]u[/tex].
[tex]\frac{x}{\sqrt{x^2+9}}[/tex] -->Because [tex]\sin(u)=\frac{\text{opp}}{\text{hyp}}[/tex]
