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

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Answer 1 · 2015-10-17T16:16:56

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

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 · 2016-03-15T03:17:52

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

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)#