Question 942007
Identity is pretty simple to verify; however you have to notice that *[tex \large \sin^4 x - 2 \sin^2 x + 1 = (\sin^2 x - 1)^2]. Using *[tex \large \sin^2 x + \cos^2 x = 1], we have *[tex \large \sin^2 x - 1 = -\cos^2 x], so the LHS equals *[tex \large (-\cos^2 x)^2 \cos x = \cos^5 x].