If you're looking for what the half angle of the tangent of theta is, I'm a bit confused as to why you think the angle in the 4th quadrant, x, is relevant. But maybe you don't know it isn't and it's a "trick" to throw you off. Hmm...

Anyways, the half angle identity for tangent is

$tan(\frac{\theta}{2})=\frac{sin\theta}{1+cos\theta}$

There are actually 3 identities for the tangent of a half angle, but this one works just as well as either of the others do, so I'm going with this one.

If theta is in QIII, the value of -4 goes along the x axis and the hypotenuse is 5. That makes the missing side, by Pythagorean's Theorem, -3. Filling in our formula:

$tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{1+(-\frac{4}{5}) }$ which simplifies a bit to

$tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{\frac{5}{5} -\frac{4}{5} }$ and a bit more to

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

Bring up the lower fraction and flip it to divide to get

$tan(\frac{\theta}{2})=-\frac{3}{5}*\frac{5}{1}$ which of course simplifies to

-3. Choice A.

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3 years ago