Question

How do you simplify `f(theta)=tan(theta/2)+3cos(theta/4)-cos(theta/4)` to trigonometric functions of a unit `theta`?

Answer 1

From half-angle formula we know that

`cos(theta/2)=sqrt[(1+costheta)/2]`

hence

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

and

#tan(theta/2)=[sin(theta/2)]/[cos(theta/2)]=> tan(theta/2)=[sqrt(1-cos^2(theta/2))]/[cos(theta/2)]=> tan(theta/2)=[sqrt(1-[(1+costheta)/2])]/[sqrt[(1+costheta)/2]]#

Hence the function becomes

#f(theta)=tan(theta/2)+2cos(theta/4)=> f(theta)=[sqrt(1-[(1+costheta)/2])]/[sqrt[(1+costheta)/2]]+2*[sqrt[(1+sqrt[(1+costheta)/2])/2]]#

Now everything is expressed as functions of unit `theta`