How do you simplify #sin4theta-cot2theta# to trigonometric functions of a unit #theta#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Shwetank Mauria Jun 23, 2016 #sin4theta-cot2theta# = #4sinthetacos^3theta-4sin^3thetacostheta-1/2(cottheta-tantheta)# Explanation: we use #sin2A=2sinAcosA# and #cos2A=cos^2A-sin^2A# Hence #sin4theta-cot2theta=2sin2thetacos2theta-(cos2theta)/(sin2theta)# = #4sinthetacostheta(cos^2theta-sin^2theta)-(cos^2theta-sin^2theta)/(2sinthetacostheta)# = #4sinthetacos^3theta-4sin^3thetacostheta-(costheta)/(2sintheta)+(sintheta)/(2costheta)# = #4sinthetacos^3theta-4sin^3thetacostheta-1/2(cottheta-tantheta)# 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 1258 views around the world You can reuse this answer Creative Commons License