How do you use a double angle formula to rewrite the expression #(cosx+sinx)(cosx-sinx)#? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer Shwetank Mauria Mar 3, 2017 #(cosx+sinx)(cosx-sinx)=cos2x# Explanation: Recall #cos2A=2cos^2A-1=1-2sin^2A=cos^2A-sin^2A# or #cos(A+B)=cosAcosB-sinAsinB#. Now as #(a+b)(a-b)=a^2-b^2# #(cosx+sinx)(cosx-sinx)# = #cos^2x-sin^2x# = #cosxcosx-sinxsinx# = #cos(x+x)# = #cos2x# 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 6015 views around the world You can reuse this answer Creative Commons License