How do you find the exact value of cos2x given #cosx=-2/13#, #pi/2<x<pi#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Nghi N. May 21, 2015 #cos 2x = 2cos^2 x - 1 = 8/169 - 1 = -161/169 = -0.95# 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 8279 views around the world You can reuse this answer Creative Commons License