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

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Answer 1 · 2018-07-22T14:16:48

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

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

# -sqrt(1/2(1+sqrt(1/2(1 + cos theta )))#.

#f(theta) = f_1 + f_2 + f_3#, with

#f_1 = - cot (theta/2) = - cos (theta/2)/(sin(theta/2))#

#= +- sqrt((1 + cos theta )/(1 - cos theta ))#,

prefixing + for #theta in Q_4#,

#f_2 = 3 csc (theta/4) = 3/sin (theta/4)= 3sqrt(2/( 1 - cos (theta/2))#

#= 3 sqrt (2/((1- sqrt(1/2( 1 + cos theta))))# and

#f_3 = - cos (theta/4) = -sqrt(1/2(1+cos (theta/2))#

#= -sqrt(1/2(1+sqrt(1/2(1 + cos theta )))#.

And so,

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

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

# -sqrt(1/2(1+sqrt(1/2(1 + cos theta )))#.