How do you prove #1 - 2 sin^2r + sin^4r = cos^4r#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Jim H May 21, 2015 Use the fact that #1+2x^2 +x^4# can be factored. It is a perfect square. #1+2x^2 +x^4 = (1-x^2)^2#. So, #1-2sin^2r+sin^4r = (1-sin^2r)^2 = (cos^2 r )^2 = cos4r# 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 2365 views around the world You can reuse this answer Creative Commons License