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

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Answer 1 · 2017-09-01T04:30:56

#cos(theta/2)-tan(theta/2)+sin(theta/2)#

= #sqrt(1+costheta)-sqrt((1-costheta)/(1+costheta))+sqrt(1-costheta)#

Recall #cos2A=2cos^2A-1=1-2sin^2A#

if #A=theta/2#. we have

#costheta=2cos^2(theta/2)-1#

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

also #2sin^2(theta/2)=-costheta#

or #sin(theta/2)=+-sqrt(1-costheta)# ........(2)

and dividing (2) by (1), we get

#tan(theta/2)=+-sqrt((1-costheta)/(1+costheta))#

We use only positive sign for #cos(theta/2),tan(theta/2),sin(theta/2)#

for working out the desired result, although multiple values of #f(theta)# are possible and we get

#cos(theta/2)-tan(theta/2)+sin(theta/2)#

= #sqrt(1+costheta)-sqrt((1-costheta)/(1+costheta))+sqrt(1-costheta)#