# How do you find the solutions to sin^-1x=tan^-1x?

Aug 19, 2016

I got $x = 0$.

$\arcsin \left(0\right) = 0$

$\arctan \left(0\right) = 0$

$f \left(x\right) = \arcsin x - \arctan x$:
graph{arcsinx - arctanx [-3.08, 3.08, -1.54, 1.54]}

This is something where you can recall that $\arcsin x$ and $\arctan x$ are actually angles. Therefore, let us choose to subtract $\arctan x$ over, and take the $\sin$ of both sides.

$\sin \left(\arcsin x - \arctan x\right) = \sin 0$

Then, we know that $\sin \left(u - v\right) = \sin u \cos v - \cos u \sin v$. Therefore:

$\implies \sin \left(\arcsin x\right) \cos \left(\arctan x\right) - \cos \left(\arcsin x\right) \sin \left(\arctan x\right) = \sin 0$

$x \cos \left(\arctan x\right) - \cos \left(\arcsin x\right) \sin \left(\arctan x\right) = 0$

Since $\arctan x$ is the angle opposite from the side of length $x$ and adjacent to the side of length $1$:

$\textcolor{g r e e n}{\cos \left(\arctan x\right) = \frac{1}{\sqrt{1 + {x}^{2}}}}$

Next, $\arcsin x$ is the angle opposite to the side of length $x$, while the hypotenuse is of length $1$. That means:

${a}^{2} + {x}^{2} = 1$

$a = \sqrt{1 - {x}^{2}}$

That means:

$\textcolor{g r e e n}{\cos \left(\arcsin x\right)} = \frac{\sqrt{1 - {x}^{2}}}{1} = \textcolor{g r e e n}{\sqrt{1 - {x}^{2}}}$

Lastly, $\sin \left(\arctan x\right)$ is the $\sin$ that uses the sides of length $x$ (opposite) and $\sqrt{1 + {x}^{2}}$ (hypotenuse), so:

$\textcolor{g r e e n}{\sin \left(\arctan x\right) = \frac{x}{\sqrt{1 + {x}^{2}}}}$

So, we can substitute these results to get:

$\frac{x}{\sqrt{1 + {x}^{2}}} - \frac{x \sqrt{1 - {x}^{2}}}{\sqrt{1 + {x}^{2}}} = 0$

$\frac{x - x \sqrt{1 - {x}^{2}}}{\sqrt{1 + {x}^{2}}} = 0$

Now let's solve for $x$.

$x - x \sqrt{1 - {x}^{2}} = 0$

$x = x \sqrt{1 - {x}^{2}}$

$1 = \sqrt{1 - {x}^{2}}$

$1 = 1 - {x}^{2}$

${x}^{2} = 0 \implies \textcolor{b l u e}{x = 0}$

This actually happens to be at the inflection point of $\arcsin x$ and $\arctan x$, and these functions are odd functions, so there are no other solutions.

$f \left(x\right) = \arcsin x - \arctan x$:
graph{arcsinx - arctanx [-3.08, 3.08, -1.54, 1.54]}

Aug 20, 2016

First of all, notice that both $\arcsin \left(x\right)$ and $\arctan \left(x\right)$ are, by definition in the interval $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

Apply function $\tan$ to our equation:
tan(arcsin(x)) = tan(arctan(x)

As we know,
$\tan \left(\phi\right) = \sin \frac{\phi}{\cos} \left(\phi\right)$ by definition of function $\tan \left(\right)$.
$\tan \left(\arctan \left(x\right)\right) = x$ by definition of $\arctan \left(\right)$.
$\sin \left(\arcsin \left(x\right)\right) = x$ by definition of $\arcsin \left(\right)$.

Therefore, our equation takes form

$\sin \frac{\arcsin \left(x\right)}{\cos} \left(\arcsin \left(x\right)\right) = x$
$\implies \frac{x}{\cos} \left(\arcsin \left(x\right)\right) = x$
Before reducing by $x$ we have to check if $x = 0$ is a solution.
Indeed it is:
$\arcsin \left(0\right) = 0$ and $\arctan \left(0\right) = 0$

So, we found one solution: $x = 0$.

Now let's reduce our equation by $x$, which, supposedly, not equal to $0$ since we already found this solution.

$\implies \frac{1}{\cos} \left(\arcsin \left(x\right)\right) = 1$
$\implies \cos \left(\arcsin \left(x\right)\right) = 1$
$\implies \arcsin \left(x\right) = 0$ since this is the only angle within $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, $\cos$ of which is $1$.

Apply function $\sin$ to both sides:
$\sin \left(\arcsin \left(x\right)\right) = \sin \left(0\right)$
$\implies x = 0$ - a solution, which we have already found and checked.

So, $x = 0$ is the only solution.