Question

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

Answer 1

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

Explanation:

`f(theta)=sin(theta/2)+cos(-theta/2-pi)`

= `sin(theta/2)+cos(-(pi+theta/2))`

= `sin(theta/2)+cos(pi+theta/2)`

= `sin(theta/2)-cos(theta/2)`

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