How do you simplify #f(theta)=2cot(theta/2)-3sin(theta/4+pi/2)# to trigonometric functions of a unit #theta#?

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Answer 1 · 2018-08-09T00:35:55

# +- 2sqrt(( 1 + cos theta )/( 1 - cos theta ))#
#- 3/sqrt2 sqrt ( 1 +- 1/sqrt2sqrt( 1 + cos theta))#. See explanation on #+_# dilemma.

As an advocate of #theta >= 0#, like #r >= 0#, I assume that #theta# is non-

negative.
#2 cot (theta/2) - 3 sin ( pi/2 + theta/4 )#

#= 2 cos ( theta/2 )/sin ( theta/2 ) - 3 cos ( theta/4 )#

#= +- 2sqrt(( 1 + cos theta )/( 1 - cos theta ))#

#- 3/sqrt2 sqrt ( 1 + cos (theta/2))#

#= +- 2sqrt(( 1 + cos theta )/( 1 - cos theta ))#

#- 3/sqrt2 sqrt ( 1 +- 1/sqrt2sqrt( 1 + cos theta))#, choosing #+#

prefix for the first term, when #theta in Q_4# and, for the same

reason, #-# in the 3rd term, under the radical sign.