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

1 Answer
Aug 8, 2018

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#