Question

How do you simplify `sin(tan^-1(x))`?

Answer 1

`sin(arctan(x)) = |x|/sqrt(x^2+1)`

Explanation:

Knowing that

`sin^2(theta) + cos^2(theta) = 1`

We divide both sides by `sin^2(theta)` so we have

`1 + cot^2(theta) = csc^2(theta)`

Or,

`1 + 1/tan^2(theta) = 1/sin^2(theta)`

Taking the least common multiple we have

`(tan^2(theta) + 1)/tan^2(theta) = 1/sin^2(theta)`

Inverting both sides we have

`sin^2(theta) = tan^2(theta)/(tan^2(theta) + 1)`

So we say that `theta = arctan(x)`

`sin^2(arctan(x)) = tan^2(arctan(x))/(tan^2(arctan(x)) + 1)`

Knowing that `tan(arctan(x)) = x`

`sin^2(arctan(x)) = x^2/(x^2 + 1)`

So we take the square root of both sides

`sin(arctan(x)) = +-sqrt(x^2/(x^2+1)) = +-|x|/sqrt(x^2+1)`

Checking the range of the arctangent, we see that during it the sine is always positive so we have

`sin(arctan(x)) = |x|/sqrt(x^2+1)`

Answer 2

`sin(tan^-1(x))=x/sqrt(x^2+1)`

Explanation:

We can use the principles of "SOH-CAH-TOA".

First, let's call `sin(tan^-1(x))=sin(theta)` where the angle `theta=tan^-1(x)`.

More specifically, `tan^-1(x)=theta` is the angle when `tan(theta)=x`. We know this from the definition of inverse functions.

Since `tan(theta)="opposite"/"adjacent"`, and here `tan(theta)=x/1` we know that

`{("opposite"=x),("adjacent"=1),("hypotenuse"=?):}`

Using the Pythagorean Theorem, we can see that the hypotenuse of a right triangle with legs `x` and `1` has `"hypotenuse"=sqrt(x^2+1)`.

Now, to find `sin(tan^-1(x))`, find `sin(theta)` for the triangle where

`{("opposite"=x),("adjacent"=1),("hypotenuse"=sqrt(x^2+1)):}`

Since `sin(theta)="opposite"/"hypotenuse"`, we see that

`sin(tan^-1(x))=x/sqrt(x^2+1)`