If #sinx=2/4#, how do you find #cos2x#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Shwetank Mauria Jul 1, 2016 #cos2x=1/2# Explanation: We have the identity #cos2x=1-2sin^2x# Hence as #sinx=2/4#, #cos2x=1-2sin^2x=1-2(2/4)^2# = #1-(2xx2xx2)/(4xx4)# = #1-1/2=1/2# 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 9030 views around the world You can reuse this answer Creative Commons License