Question

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

Answer 1

I got `x = 0`.

`arcsin(0) = 0`

`arctan(0) = 0`

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

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

`sin(arcsinx - arctanx) = sin0`

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

`=> sin(arcsinx)cos(arctanx) - cos(arcsinx)sin(arctanx) = sin0`

`xcos(arctanx) - cos(arcsinx)sin(arctanx) = 0`

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

`color(green)(cos(arctanx) = 1/(sqrt(1+x^2)))`

Next, `arcsinx` 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:

`color(green)(cos(arcsinx)) = sqrt(1 - x^2)/1 = color(green)(sqrt(1 - x^2))`

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

`color(green)(sin(arctanx) = x/(sqrt(1 + x^2)))`

So, we can substitute these results to get:

`x/(sqrt(1+x^2)) - (xsqrt(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 - xsqrt(1 - x^2) = 0`

`x = xsqrt(1 - x^2)`

`1 = sqrt(1 - x^2)`

`1 = 1 - x^2`

`x^2 = 0 => color(blue)(x = 0)`

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

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

Answer 2

First of all, notice that both `arcsin(x)` and `arctan(x)` are, by definition in the interval `[-pi/2,pi/2]`.

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

As we know,
`tan(phi)=sin(phi)/cos(phi)` by definition of function `tan()`.
`tan(arctan(x))=x` by definition of `arctan()`.
`sin(arcsin(x)) = x` by definition of `arcsin()`.

Therefore, our equation takes form

`sin(arcsin(x))/cos(arcsin(x)) = x`
`=> x/cos(arcsin(x)) = x`
Before reducing by `x` we have to check if `x=0` is a solution.
Indeed it is:
`arcsin(0) = 0` and `arctan(0) = 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.

`=> 1/cos(arcsin(x)) = 1`
`=> cos(arcsin(x)) = 1`
`=> arcsin(x) = 0` since this is the only angle within `[-pi/2,pi/2]`, `cos` of which is `1`.

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

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