Solution

Recall from chapter 6 that any Inverse Trig algebraic statement can be expressed as the angle of some right triangle. Thus, let's say that \(\theta = \tan^{-1}(x)\) and then draw the corresponding right triangle:

Depsite having drawn this in the 4th quadrant, you can choose to draw it in the 1st quadrant as well:

In either case, we need to solve for the hypotenuse: \[ \solve{ H^2 &=& x^2+1\\ H &=& \sqrt{x^2+1} } \]

Now that we have the hypotenuse, we can come back to the original statement: \[ \sin(2\tan^{-1}(x)) = \sin(2\theta) \]Where \(\theta\) is the angle as drawn in our triangle above. To find the double angle for Sine, we need both the Sine and Cosine of this triangle: \[ \solve{ \sin(\theta) &=& \frac{{x}}{\sqrt{x^2+1} }\\ \cos(\theta) &=& \frac{{1}}{\sqrt{x^2+1} } }\]So that when we apply the double sine angle formula: \[ \solve{ \sin(2\theta) &=&2\sin(\theta)\cos(\theta)\\ &=&2\left(\frac{{x}}{\sqrt{x^2+1} }\right)\left(\frac{{1}}{\sqrt{x^2+1} }\right)\\ \sin(2\theta)&=&\dfrac{ 2x }{ x^2 +1 } } \]

Thus, our answer can finally be expressed: \[ \sin(2\tan^{-1}(x))=\sin(2\theta) = \dfrac{{2x}}{ x^2 +1} \]