Prove the identity of #cos(3x + 3y) + cos(3x − 3y) = 2 cos 3x cos 3y#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer P dilip_k Dec 4, 2016 using identities #cos(A+B)=cosAcosB-sinAsinB# #cos(A-B)=cosAcosB+sinAsinB# weget #LHS=cos(3x + 3y) + cos(3x − 3y) # #=cos3xcos3y-cancel(sin3xsin3y)+cos3xcos3y+cancel(sin3xsin3y)# #=2cos3xcos3y=RHS# Proved Answer link Related questions What does it mean to prove a trigonometric identity? How do you prove #\csc \theta \times \tan \theta = \sec \theta#? How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#? How do you prove that #cos 2x(1 + tan 2x) = 1#? How do you prove that #(2sinx)/[secx(cos4x-sin4x)]=tan2x#? How do you verify the identity: #-cotx =(sin3x+sinx)/(cos3x-cosx)#? How do you prove that #(tanx+cosx)/(1+sinx)=secx#? How do you prove the identity #(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)#? See all questions in Proving Identities Impact of this question 2104 views around the world You can reuse this answer Creative Commons License