Question #cf8dc Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Nghi N Apr 4, 2017 #(sin x + cos x)^2 = sin^2 x + cos^2 x + 2sin x.cos x # Since, #sin^2 x + cos^2 x = 1# #2sin x.cos x = sin 2x#, Therefore, #(sin x + cos x)^2 = 1 + sin 2x# 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 943 views around the world You can reuse this answer Creative Commons License