Hmm, let's derive the half angle formulas, which come from the cosine double angle formulas:
#cos(2x) = cos^2 x - sin ^ 2 x = 2 cos^2 x - 1 = 1 - 2 sin ^2 x#
#cos x = pm sqrt{ 1 / 2 (1 + cos(2x) ) } #
#sin x = pm sqrt{1/2 (1 - cos(2x) ) } #
Let's set #x=theta/2#
#cos(theta/2) = pm sqrt{ 1 / 2 (1 + cos theta ) } #
#sin(theta/2) = pm sqrt{ 1 / 2 (1 - cos theta ) } #
Let's set #x=theta/4#
#cos(theta/4) = pm sqrt{1/2( 1+ cos(theta/2)) } #
# = pm sqrt{1/2( 1pm sqrt{ 1 / 2 (1 + cos theta ) } ) } #
#sin(theta/4) = pm sqrt{1/2( 1pm sqrt{ 1 / 2 (1 - cos theta ) } ) } #
# f(theta) = 2 tan (theta/2) - sin (theta/4) - sec(theta/4) #
# f(theta) = { pm 2 sqrt{ 1 / 2 (1 - cos theta ) } }/ { pm sqrt{ 1 / 2 (1 + cos theta ) } } pm sqrt{1/2( 1pm sqrt{ 1 / 2 (1 - cos theta ) } ) } - 1/{pm sqrt{1/2( 1+ cos(theta/2)) } }
#
# f(theta) = pm 2 sqrt{ {1 - cos theta }/ { 1 + cos theta } } pm sqrt{1/2( 1pm sqrt{ 1 / 2 (1 - cos theta ) } ) } - 1/{pm sqrt{1/2( 1+ cos(theta/2)) } }
#
That doesn't seem particularly simplified to me, but there it is.
