How do you simplify sin⁡(tan−1(x))?

Question

Paul Mitchell · Accepted Answer

Explanation:  Knowing that  sin2(θ)+cos2(θ)=1  We divide both sides by sin2(θ) so we have  1+cot2(θ)=csc2(θ)  Or,  1+1tan2(θ)=1sin2(θ)  Taking the least common multiple we have  tan2(θ)+1tan2(θ)=1sin2(θ}  Inverting both sides we have  sin2(θ)=tan2(θ)tan2(θ)+1  So we say that θ=arctan⁡(x)  sin2(arctan⁡(x))=tan2(arctan⁡(x))tan2(arctan⁡(x))+1  Knowing that tan⁡(arctan⁡(x))=x  sin2(arctan⁡(x))=x2x2+1  So we take the square root of both sides  sin⁡(arctan⁡(x))=±x2x2+1=±|x|x2+1

Deufemiak7 · Accepted Answer

We can use the principles of &quot;SOH-CAH-TOA&quot;. First, let&#039;s call sin⁡(tan−1(x))=sin⁡(θ) where the angle θ=tan−1(x) More specifically, tan−1(x)=θ is the angle when tan⁡(θ)=x. We know this from the definition of inverse functions. Since tan⁡(θ)=oppositeadjacent, and here tan⁡(θ)=x1 we know that {Opposite=xadjacent=1hypotenuse=? Using the Pythagorean Theorem, we can see that the hypotenuse of a right triangle with legs x and 1 has hypotenuse =x2+1 Now find sin⁡(tan−1(x)) Since sin⁡(θ)=oppositehypotenuse, we see that sin⁡(tan−1(x))=xx2+1

karton · Accepted Answer

Finaly find answer. Thank you.

How do you simplify \sin(\tan^{-1}(x))?

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