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Say I have the trig identity

$$ \tan \frac{\theta}{2} = \frac{ 1 - \cos \theta }{ \sin \theta } $$

And you have the 3,4,5 triangle:

taking the angle $\theta$ as marked, why can't we just say

$$ \tan \frac{\theta}{2} = \frac{ 1.5 }{ 4 } = \frac{ 3 }{ 8 } $$

(Like, if you half $\theta$, then the opposite side length is seen as half, right?)

But this doesn't work or check out with the identity:

$$ \tan \frac{\theta}{2} = \frac{ 1 - \frac{4}{5} }{ \frac{3}{5} } = \frac{ \frac{1}{5} }{ \frac{3}{5} } = \frac{1}{3} $$

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6  
"(Like, if you halve θ, then the opposite side length is seen as half, right?)" - nope. If you draw a line at an angle θ/2 from the "4" leg, it won't intersect the "3" leg at its midpoint. –  J. M. Apr 27 '11 at 23:58
4  
Imagine the case where $\theta$ is a large acute angle, like $89^\circ$. Then $\tan\theta$ is big, but $\tan\frac{\theta}{2}<1$ –  Jonas Meyer Apr 28 '11 at 0:14

1 Answer 1

up vote 7 down vote accepted

Actually, if you half an angle, it will divide the opposite side proportionally to the two other sides. (see http://en.wikipedia.org/wiki/Angle_bisector_theorem )

In your case, 3 would be divided into parts $4/3$ and $5/3$.

So you get $\tan \theta/2=(4/3)/4= 1/3$. Everything works out fine.

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