How do you use the half angle identify to find the exact value of #tan22.5^circ#?

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Answer 1 · 2016-11-16T13:19:06

The half angle is, #tan22.5º=(-1+sqrt2)/2#

We start from the definition of #tantheta#

#tantheta=sintheta/costheta=(2sin(theta/2)cos(theta/2))/(cos^2(theta/2)-sin^2(theta/2))#

Dividing by #cos^2(theta/2)#

#tantheta=(2(sin(theta/2))/cos(theta/2))/(1-tan^2(theta/2))#

#=(2tan(theta/2))/(1-tan^2(theta/2))#

Let #tan(theta/2)=t#

We need #tan22.5= tan(theta/2)#

#tantheta=tan45º=1#

Therefore
#1=(2t)/(1-t^2)#

#1-t^2=2t#

#t^2+2t-1=0#

We solve this quadratic equation with

#Delta=4+4=8#

#Delta>0#, therefore 2 real roots

#t=(-2+-(sqrt8))/2=(-1+-sqrt2)#

We keep the positive root

#t=(-1+sqrt2)/2 = tan22.5º#