Question

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

Answer 1

`2(2csc theta + cot theta )-3/8sqrt(8sqrt8) (sqrt ((2sqrt2 +- sqrt(1+ cos theta ))))`,
choosing `-` of `(+-)`, for `0 < theta in Q_4`.

Explanation:

`theta /4 in Q_1 and theta/2 in Q_2`,

for `0 < theta in Q_4`, and then, cos (theta/2) < 0#.

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

`= (3 cos^2(theta/2) + sin ^2(theta/2))/(sin(theta/2)cos(theta/2))`

`-3 sqrt (1/2(1 + cos(theta/2))`

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

`= 2(2csc theta + cot theta )-(3/8)sqrt(8sqrt8) (sqrt ((2sqrt2 +- sqrt(1+ cos theta ))))`,

choosing `-` of `(+-)`, for `0 < theta in Q_4`.

Note that,

if `0 < theta in Q_4 in ( - pi/2, 0 ), theta/2 in Q_4 in ( - pi/4, 0 )`

and, the choice is + ( for `cos ( theta/2 )`). Indeed, incredible.

I am not approving ( like my invalidating `r < 0` ), use of negative

`theta`. This contradiction is the reason.

As these are not in tune with the universal faith on conventions, I

do not invite any discussion. I would like to be alone as non-

conventional, on such foundation matters.