How do you prove #9sin(3x)cos(3x)#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer José F. Feb 5, 2016 #(9/2)sin(6x)# Explanation: we know that #2sin(x)cos(x)=sin(2x)# then: #9sin(3x)cos(3x)=(9/2)*2sin(3x)cos(3x)# #(9/2)sin(6x)# 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 2297 views around the world You can reuse this answer Creative Commons License