Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 63

$cos(arcsin~2x)=\sqrt {1-4x^2}$

Let $u=arcsin~2x~~$ (Range: $-\frac{\pi}{2}\leq u\leq\frac{\pi}{2}$). Then: $2x=sin~u~~$ $cos^2u+sin^2u=1$ $cos^2u=1-sin^2u$ $cos^2u=1-(2x)^2$ $cos^2u=1-4x^2$ $cos~u=+\sqrt {1-4x^2}$ $cos(arcsin~2x)=\sqrt {1-4x^2}$