How do you find the solutions to ${sin}^{- 1} x = {tan}^{- 1} x$ $sin^-1x=tan^-1x$ ?

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How do you find the solutions to ${sin}^{- 1} x = {tan}^{- 1} x$ $sin^-1x=tan^-1x$ ?

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Answer 1 · 2016-08-19T03:09:43

I got $x = 0$ $x = 0$ .

$arcsin (0) = 0$ $arcsin(0) = 0$

$arctan (0) = 0$ $arctan(0) = 0$

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

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

$sin (arcsin x - arctan x) = sin 0$ $sin(arcsinx - arctanx) = sin0$

Then, we know that $sin (u - v) = sin u cos v - cos u sin v$ $sin(u-v) = sinucosv - cosusinv$ . Therefore:

$\Rightarrow sin (arcsin x) cos (arctan x) - cos (arcsin x) sin (arctan x) = sin 0$ $=> sin(arcsinx)cos(arctanx) - cos(arcsinx)sin(arctanx) = sin0$

$x cos (arctan x) - cos (arcsin x) sin (arctan x) = 0$ $xcos(arctanx) - cos(arcsinx)sin(arctanx) = 0$

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

$cos (arctan x) = \frac{1}{\sqrt{1 + x^{2}}}$ $color(green)(cos(arctanx) = 1/(sqrt(1+x^2)))$

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

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

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

That means:

$cos (arcsin x) = \frac{\sqrt{1 - x^{2}}}{1} = \sqrt{1 - x^{2}}$ $color(green)(cos(arcsinx)) = sqrt(1 - x^2)/1 = color(green)(sqrt(1 - x^2))$

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

$sin (arctan x) = \frac{x}{\sqrt{1 + x^{2}}}$ $color(green)(sin(arctanx) = x/(sqrt(1 + x^2)))$

So, we can substitute these results to get:

$\frac{x}{\sqrt{1 + x^{2}}} - \frac{x \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}}} = 0$ $x/(sqrt(1+x^2)) - (xsqrt(1 - x^2))/(sqrt(1 + x^2)) = 0$

$\frac{x - x \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}}} = 0$ $(x - xsqrt(1 - x^2))/(sqrt(1 + x^2)) = 0$

Now let's solve for $x$ $x$ .

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

$x = x \sqrt{1 - x^{2}}$ $x = xsqrt(1 - x^2)$

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

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

$x^{2} = 0 \Rightarrow x = 0$ $x^2 = 0 => color(blue)(x = 0)$

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

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

Answer 2 · 2016-08-20T02:50:12

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

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

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

Therefore, our equation takes form

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

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

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

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

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

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

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