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

Question

1334 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-29T17:10:07

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

We use the following

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

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

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

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

#tan(theta/2)=sin(theta/2)/cos(theta/2)#

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

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

#=(1-costheta)/sintheta#

Similarly,

#cot(theta/2)=cos(theta/2)/(sin(theta/2)#

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

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

#=(1+costheta)/sin theta#

#sin(theta/4)=sqrt((1-cos(theta/2))/2)#

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

And finally,

#f(theta)=sin(theta/4)+tan(theta/2)+cot(theta/2)#

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

#=sqrt((1-sqrt((1+costheta)/2))/2)+2/sintheta#