Question

4cos2x-2cos^2x, если sin^2x=1/6

Answer 1

$4cos2x-2cos^{2}x=4( cos^{2}x-sin^{2}x)-2(1-sin^{2}x)=4cos^{2}x- \ -4sin^{2}x-2+2sin^{2}x=4(1-sin^{2}x)-4sin^{2}x-2+2sin^{2}x= \ =4-4sin^{2}x-4sin^{2}x-2+2sin^{2}x=2-6sin^{2}x$
При $sin^{2}x= frac{1}{6}$
$2-6* frac{1}{6}=2-1=1$