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

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

First, let's call

More specifically,

Since

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

Using the Pythagorean Theorem, we can see that the hypotenuse of a right triangle with legs

Now, to find

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

Since

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

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Knowing that

We divide both sides by

Or,

Taking the least common multiple we have

Inverting both sides we have

So we say that

Knowing that

So we take the square root of both sides

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

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