How do you write #2sin0.6cos0.6# as a single trigonometric function? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Shwetank Mauria Sep 27, 2016 #2sin0.6cos0.6=sin1.2# Explanation: As #sin2A=2sinAcosA# #2sin0.6cos0.6=sin(2xx0.6)=sin1.2# 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 3970 views around the world You can reuse this answer Creative Commons License