How do you prove # (cos(x/2)-sin(x/2))^2= 1-sinx#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Bdub Apr 17, 2016 See below Explanation: Use Properties: #sin^2x+cos^2x=1, and sin2x=2sinxcosx# Left Side #=cos^2(x/2)-2sin(x/2)cos(x/2)+sin^2(x/2)# #=cos^2(x/2)+sin^2(x/2)-2sin(x/2)cos(x/2)# #=1-2sin(x/2)cos(x/2)# #=1-sin(2(x/2))# #=1-sinx# #=#Right side 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 30714 views around the world You can reuse this answer Creative Commons License