First check the supposed identity by plugging in A=π/4. sin(A)=1√2, sin2(A)=½, sin4(A)=¼.
cos(4A)=cos(π)=-1.
1-sin2(A)-⅛sin4(A)=1-½-1/32≠-1, so the identity is false.
cos(4A)=1-2sin2(2A)=
1-2(2sin(A)cos(A))2=1-8sin2(A)cos2(A)=
1-8sin2(A)(1-sin2(A))=1-8sin2(A)+8sin4(A).
We know this should be -1 when A=π/4. 1-8sin2(A)+8sin4(A)=1-4+2=-1, which is correct.
When A=π/6, sin2(A)=¼, sin4(A)=1/16, cos(⅔π)=-½. 1-8sin2(A)+8sin4(A)=1-2+½=-½, which is also correct.
Therefore the identity should be cos(4A)=1-8sin2(A)+8sin4(A) and not the given supposed identity.