How do you prove #cos4x=1-2sin^2(2x)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Narad T. Apr 13, 2018 Please see the proof below Explanation: #"Reminder"# #cos(A+B)=cosAcosB-sinAsinB# #cos^2theta+sin^2theta=1# Therefore, #LHS=cos4x=cos(2x+2x)# #=cos2xcos2x-sin2xsin2x# #=cos^2 2x- sin^2 2x# #=1-sin^2 2x-sin^2 2x# #=1-2sin^2 2x# #=RHS# #QED# 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 25112 views around the world You can reuse this answer Creative Commons License