Chapter 6 - Analytic Trigonometry - Section 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 521: 102

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

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