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

1 Answer
Shiva Prakash M V
Feb 16, 2018

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

Explanation:

#f(theta/4)=cos(theta/4)+sec^2(theta/2)+cot(theta/2)#

#cos(theta/4)=cos(1/2theta/2)=sqrt((1+cos(1/2theta))/2)#
#sec^2(theta/2)=1/cos^2(theta/2)=1/((1+costheta)/2)=2/(1+costheta)#
#cot(theta/2)=cos(theta/2)/(sin(theta/2))=sqrt(1+costheta)/(sqrt(1-costheta)#
Now,
#cos(theta/4)=sqrt((1+sqrt((1+costheta)/2))/2#

Thus,
#f(theta/4)=sqrt(1+sqrt((1+costheta)/2))/2+2/(1+costheta)+sqrt(1+costheta)/(sqrt(1-costheta)#

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