Not sure what the question is, but I guess it's to prove that
[tex]\cos3A=4\cos^3A-3\cos A[/tex]
Expand the left side as
[tex]\cos3A=\cos(2A+A)=\cos2A\cos A-\sin2A\sin A[/tex]
and use the double angle identities to write
[tex]\cos3A=(\cos^2A-\sin^2A)\cos A-2\sin^2A\cos A[/tex]
[tex]\cos3A=\cos^3A-3\sin^2A\cos A[/tex]
Recall the Pythagorean identity:
[tex]\cos3A=\cos^3A-3(1-\cos^2A)\cos A[/tex]
[tex]\cos3A=\cos^3A-3\cos A+3\cos^3A[/tex]
[tex]\implies\cos3A=4\cos^3A-3\cos A[/tex]
as required.
