How do you express #f(theta)=sin(theta/2)+cos^2(theta/2)# in terms of trigonometric functions of a whole theta?

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Answer 1 · 2018-07-23T16:18:44

# sqrt( 1/2(1 - cos theta )) + 1/2( 1 + cos theta), if theta > 0#.
# - sqrt( 1/2(1 - cos theta )) + 1/2( 1 + cos theta), if theta < 0#.

For #theta in # any quadrant,

#theta/2 in Q_1# or #Q_2, if theta > 0# and.

Here, #sin(theta/2) > 0#.

#theta/2 in Q_3# or #Q_4, if theta < 0#

Here, #sin(theta/2) < 0#.

#f ( theta ) = sin (theta/2) + cos^2(theta/2)#

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

#cos 2A = cos^2A - sin^2A = 2 cos^2A - 1 = 1 - 2 sin^2A#