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

$t=(-1+sqrt2)/2 = tan22.5º$

How do you use the half angle identify to find the exact value of tan22.5^circtan22.5^circ?