How do you verify cos^4x - sin^4x = cos^2x - sin^2x? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer George C. May 22, 2015 cos^4x-sin^4x = (cos^2x)^2-(sin^2x)^2 = (cos^2x-sin^2x)(cos^2x+sin^2x) = (cos^2x-sin^2x)xx1 = (cos^2x-sin^2x) based on the identities: (a^2-b^2) = (a-b)(a+b) cos^2x+sin^2x = 1 from Pythagoras (right angled triangle with hypotenuse of length 1). Answer link Related questions What does it mean to prove a trigonometric identity? How do you prove \csc \theta \times \tan \theta = \sec \theta? How do you prove (1-\cos^2 x)(1+\cot^2 x) = 1? How do you show that 2 \sin x \cos x = \sin 2x? is true for (5pi)/6? How do you prove that sec xcot x = csc x? How do you prove that cos 2x(1 + tan 2x) = 1? How do you prove that (2sinx)/[secx(cos4x-sin4x)]=tan2x? How do you verify the identity: -cotx =(sin3x+sinx)/(cos3x-cosx)? How do you prove that (tanx+cosx)/(1+sinx)=secx? How do you prove the identity (sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)? See all questions in Proving Identities Impact of this question 18781 views around the world You can reuse this answer Creative Commons License