Can't figure out how $\sin\left(\tan^{-1}(x)\right)=\frac{x}{\sqrt {x^2+1}}$

Question

Simplify the expression $$\sin\left(\tan^{-1}(x)\right)$$ Using a triangle with an angle $\theta$, opposite is x and adjacent is 1 meaning the hypo. is ${\sqrt {x^2+1}}$

Now because the problem has sin after the $\tan^{-1}(x)$ that means that the opp. being x, is put on top of the hypo. giving us the answer, but I feel like I'm missing something because I haven't incorporated the $\tan^{-1}x$

Answer 1

Let $\tan^{-1}x=y\implies x=\tan y$ and $-\dfrac\pi2<y<\dfrac\pi2$ using Principal values

$\implies\cos y>0,\cos y=+\dfrac1{\sqrt{1+\tan^2y}}=?$

$\sin y=\cos y\cdot\tan y=?$

Answer 2

You're thinking about it just right. $\tan^{-1}(x)$ is the angle whose tangent is $x$. So draw a right triangle and put $\tan^{-1}(x)$ in a corner angle. Since the tangent of that angle is $x$, the opposite side divided by the adjacent side must be $x$: the easiest choice is $x$ (opposite side) and $1$ (adjacent side) (though you could choose $1$ and $1/x$, or $\sqrt{x}$ and $1/\sqrt{x}$, or whatever you like with the right ratio). Then the hypotenuse is $\sqrt{1+x^2}$, and the sine of your angle, being the opposite side divided by the hypotenuse, is $x/\sqrt{1+x^2}$.

Answer 3

A solution with differential equations: let $f(x):=\sin(\tan^{-1} (x))$ and $g(x):= \frac{x}{\sqrt {x^2+1}}.$

Then it is easy to see that $f$ and $g$ are solutions of the second order intial value problem

$$(1+x^2)^2y''+2x(1+x^2)y'+y=0, \quad y(0)=0, \quad y'(0)=1$$

on $ \mathbb R.$

But this intial value problem has a unique solution on $ \mathbb R, $ thus $f=g$ on $ \mathbb R.$