If #sinx=(4/5)#, how do you find #sin2x#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Yonas Yohannes Apr 11, 2016 #sin2x = 2*4/5*3/5=24/25# Explanation: Since #sinx=4/5# we have a (3, 4, 5) triangle and we can totally define all sines and cosines, #cosx=3/5# Now we use the double angle identities: #sin2x=2sinx*cosx# #sin2x = 2*4/5*3/5=24/25# 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 44660 views around the world You can reuse this answer Creative Commons License