How do you prove #cos2x = 2cos^2x -1#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Nghi N. May 24, 2015 Use the identity: cos (a + b) = cos a.cos b - sin a.sin b #cos 2x = cos (x + x) = cos x.cos x - sin x. sin x = cos^2 x - sin^2 x #= #= cos^2 x - (1 - cos^2 x) = 2cos ^2 x - 1 # Answer link Related questions What are Double Angle Identities? How do you use a double angle identity to find the exact value of each expression? How do you use a double-angle identity to find the exact value of sin 120°? How do you use double angle identities to solve equations? How do you find all solutions for #sin 2x = cos x# for the interval #[0,2pi]#? How do you find all solutions for #4sinthetacostheta=sqrt(3)# for the interval #[0,2pi]#? How do you simplify #cosx(2sinx + cosx)-sin^2x#? If #tan x = 0.3#, then how do you find tan 2x? If #sin x= 5/3#, what is the sin 2x equal to? How do you prove #cos2A = 2cos^2 A - 1#? See all questions in Double Angle Identities Impact of this question 120512 views around the world You can reuse this answer Creative Commons License