Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 68: 72

$sin (2~arcos~x)= 2x~\sqrt{1-x^2}$

Suppose that: $cos ~\theta = x$ Then: $\frac{adj}{hyp} = \frac{x}{1}$ Then: $opp = \sqrt{1-x^2}$ We can find $sin (2 ~arccos ~x)$: $sin (2~arcos~x)$ $= sin (2~\theta)$ $= 2~sin (\theta)~cos(\theta)$ $= 2~\cdot \frac{opp}{hyp}~\cdot \frac{adj}{hyp}$ $= 2\cdot \frac{\sqrt{1-x^2}}{1}\cdot \frac{x}{1}$ $= 2x~\sqrt{1-x^2}$