How do you verify #sin^5xcos^2x=(cos^2x-2cos^4x+cos^6x)sinx#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Cesareo R. Jul 3, 2016 See below Explanation: Operating in both sides #sin^5xcos^2x=(cos^2x-2cos^4x+cos^6x)sinx# #sin xsin^4xcos^2x=(cos^2x-2cos^4x+cos^6x)sin x# #sin xsin^4xcos^2x=(1-2cos^2x+cos^4x)cos^2x sin x# #sin xsin^4xcos^2x=(1-cos^2x)^2cos^2x sin x# #sin xsin^4xcos^2x=sin^4xcos^2x sin x# 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 10994 views around the world You can reuse this answer Creative Commons License