\(1-\frac{1}{4}sin^2x+cosx=1-\frac{1}{4}\left(1-cos^2x\right)+cosx\)
\(=\frac{3}{4}+\frac{1}{4}cos^2x+cosx=\frac{3}{4}+\frac{1}{4}\left(2cos^2\frac{x}{2}-1\right)^2+2cos^2\frac{x}{2}-1\)
\(=\frac{1}{4}\left(4cos^4\frac{x}{2}-4cos^2\frac{x}{2}+1\right)+2cos^2\frac{x}{2}-\frac{1}{4}\)
\(=cos^4\frac{x}{2}+cos^2\frac{x}{2}\)