Answer
The identity is verified.
$sin^5x~cos^2x=(cos^2x-2~cos^4x+cos^6x)sin~x$
Work Step by Step
$sin^2x+cos^2x=1$ (Pythagorean identity)
$sin^2x=1-cos^2x$
$sin^5x~cos^2x=sin~x~sin^4x~cos^2x=sin~x~(sin^2x)^2~cos^2x=sin~x~(1-cos^2x)^2~cos^2x=sin~x~(1-2~cos^2x+cos^4x)~cos^2x=sin~x~(cos^2x-2~cos^4x+cos^6x)=(cos^2x-2~cos^4x+cos^6x)sin~x$
