[tex]\cos^3x=\cos^2x\cdot\cos x=\dfrac{1+\cos2x}2\cdot\cos x=\dfrac12\cos x+\dfrac12\cos2x\cos x[/tex]
Recall that
[tex]\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta[/tex]
from which we can find that
[tex]\cos\alpha\cos\beta=\dfrac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}2[/tex]
Then
[tex]\cos2x\cos x=\dfrac{\cos3x+\cos x}2[/tex]
and we end up with
[tex]\cos^3x=\dfrac34\cos x+\dfrac14\cos3x[/tex]
