Question

How do you express `f(theta)=cos(theta/2)-tan(theta/4)+sec(theta/2)` in terms of trigonometric functions of a whole theta?

Answer 1

`f(theta)=((3+cos theta)*sqrt(2+2 cos theta))/(2+2 cos theta)+(sin theta-sqrt(2-2 cos theta))/(1-cos theta)`

Explanation:

Let us begin with `cos (theta/2)`

`cos (theta/2)=sqrt((1+cos theta)/2)` Half-angle formula

Next, the `Sec (theta/2)` is the reciprocal of `cos (theta/2)`

That `sec (theta/2)=sqrt(2/(1+cos theta))`

Next, the `tan (theta/4)` can be expressed using sine and cosine:

`tan (theta/4)=sqrt((1-cos (theta/2))/2)/(sqrt((1+cos (theta/2))/2))=(1-cos(theta/2))/sin (theta/2)`
`tan (theta/4)=csc(theta/2)-cot(theta/2)`

Also `cot(theta/2)=sin theta/(1-cos theta)` Half-angle formula for cotangent.

We can use `theta` now

`tan (theta/4)=sqrt(2/(1-cos theta))-sin theta/(1-cos theta)`
Combine the three expressions
`f(theta)=sqrt((1+cos theta)/2)-(sqrt(2/(1-cos theta))-sin theta/(1-cos theta))+sqrt(2/(1+cos theta))`

Final answer after simplification

`f(theta)=((3+cos theta)*sqrt(2+2 cos theta))/(2+2 cos theta)+(sin theta-sqrt(2-2 cos theta))/(1-cos theta)`

Have a nice day from the Philippines !!!!!