How do you simplify #cosx(2sinx + cosx)-sin^2x#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Gió Feb 16, 2015 You can start by multiplying #cos(x)# to get: #2sin(x)cos(x)+cos^2(x)-sin^2(x)=# or: #=[2sin(x)cos(x)]+[cos^2(x)-sin^2(x)]=# which can be written as: #=sin(2x)+cos(2x)# 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]#? 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#? How can tan 4x be simplified or sec 2x? See all questions in Double Angle Identities Impact of this question 6332 views around the world You can reuse this answer Creative Commons License