How do you use the trig identity #cos2x = cos^2 x - sin^2 x # to verify that #cos 2x = 2 cos^2 x -1#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer sente Jan 20, 2016 From the identity #sin^2(x) + cos^2(x) = 1# we can subtract #cos^2(x)# to obtain #sin^2(x) = 1-cos^2(x)#. Then, #cos(2x) = cos^2(x) - sin^2(x)# #= cos^2(x) - (1 - cos^2(x))# #= 2cos^2(x) - 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 13042 views around the world You can reuse this answer Creative Commons License