Question

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

Answer 1

Answer is in the explanation.

Explanation:

It is my rule that, on par with `r >= 0, theta >= 0`

In other words, for Australians, the Earth spins clockwise and, for

Mexicans, it is anticlockwise.

So, both can keep the spin `theta >= 0`. This is universal relativity,

with respect to the advancing real time. And so,

if `theta in Q_4, theta/2 in Q_2` and

if `theta in Q_3, theta/2 in Q_1`.

For any `theta, theta/4 in Q_1`. Now,

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

`- 4sqrt2/( 1+ cos (theta/2)) + 1/sqrt2 sqrt( 1 - cos (theta/2) )`

`= +- sqrt (( 1 - cos theta )/( 1 + cos theta ))`

`- 4sqrt2 /sqrt (( 1 + sqrt(1/ 2( 1+ cos theta )))`

`+ 1/sqrt2 sqrt( 1 - sqrt( 1/2( 1 + cos theta) )`.

Choose the prefix sign `-`, for the 1st term,

when `0 <= theta in Q_4`.

Example: `theta = 330^o`, `theta/2= 165^o in Q_2`,

wherein `tan (theta/2) = tan 165^o = - 0.2679... < 0`