Chapter 6 - Analytic Trigonometry - Section 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 521: 101

$\frac{4x}{x^2+4}$

1. Given $x=2tan\theta$, we have $tan\theta=\frac{x}{2}, sin\theta=\frac{x}{\sqrt {x^2+4}}, cos\theta=\frac{2}{\sqrt {x^2+4}}$, 2. We have $sin(2\theta)=2sin\theta cos\theta=2(\frac{x}{\sqrt {x^2+4}})(\frac{2}{\sqrt {x^2+4}})=\frac{4x}{x^2+4}$