What is #tan(theta/2)# in terms of trigonometric functions of a unit #theta#?

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Answer 1 · 2016-03-09T13:28:48

#tan(theta/2)=(-1+-sectheta)/(tantheta)#

We will use the identity #tantheta=(2tan(theta/2))/(1-tan^2(theta/2)#.

Let #x=tan(theta/2)# then

#tantheta=(2x)/(1-x^2)# or

#tantheta(1-x^2)=2x# or #-tanthetax^2-2x+tantheta=0# or

#tanthetax^2+2x-tantheta=0#.

Now using quadratic formula

#x=(-2+-sqrt(2^2-4xxtanthetaxx(-tantheta)))/(2tantheta)#

#x=(-2+-sqrt(4+4tan^2theta))/(2tantheta)# or

#x=(-2+-2sqrt(sec^2theta))/(2tantheta)# or

#x=(-2+-2sectheta)/(2tantheta)#

#x=(-1+-sectheta)/(tantheta)# or

#tan(theta/2)=(-1+-sectheta)/(tantheta)#