How do you express #f(theta)=cos(theta/4)+csc(theta/2)+sin^2(theta/2)# in terms of trigonometric functions of a whole theta?

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Answer 1 · 2016-11-28T16:21:17

The answer is #=sqrt(1/2(1+sqrt((1+costheta)/2)))+sqrt(2/(1-costheta))+(1-costheta)/2#

We are going to use,

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

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

#csc(theta/2)=1/sin(theta/2)=sqrt(2/(1-costheta))#

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

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

and #cos(theta/2)=2cos^2(theta/4)-1#

#cos^2(theta/4)=(1+cos(theta/2))/2=1/2(1+sqrt((1+costheta)/2))#

#cos(theta/4)=sqrt(1/2(1+sqrt((1+costheta)/2)))#

#f(theta)=cos(theta/4)+ csc(theta/2)+sin^2(theta/2)#

And finally,

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