Algebra and Trigonometry 10th Edition

$cos^2x-\frac{1}{2}=\frac{1}{2}~cos~2x$
$sin^2x+cos^2x=1$ $cos^2x-1=-sin^2x$ Use: $cos~2x=cos^2x-sin^2x=2~cos^2x-1$ $cos^2x-\frac{1}{2}=\frac{1}{2}(2~cos^2x-1)=\frac{1}{2}~cos~2x$