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

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 $$\tan^{-1}x$$

• Have you drawn the picture? – Lubin Jan 18 at 6:09

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

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

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

• where'd you get the cos y from? – Eric Brown Jan 18 at 6:04
• @EricBrown, $$\sec^2y-\tan^2y=?$$ – lab bhattacharjee Jan 18 at 6:12

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 $$x/\sqrt{1+x^2}$$.

• Right. Just how I’d do it. – Lubin Jan 18 at 6:09
• So replace $\theta$ with $\tan^{-1}(x)$? – Eric Brown Jan 18 at 6:11

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 $$\mathbb R.$$