Unlike others I've tried, I'm having a hard time with this half-angle exercise:
If $tan(\theta)={3}$ and $\theta$ is in QIII, find $\tan\left(\frac{\theta}{2}\right)$
Here's what I know (or think I know):
- $\cos(\theta)=\left(-\frac{\sqrt{10}}{10}\right)$
- I shold use the half-angle formula for tan: $\tan\left(\frac{\theta}{2}\right)$ $\pm\sqrt\frac{1-\cos\theta}{1+\cos\theta}$
- I know the answer is:
$$\frac{\sqrt{10}+1}{-3}$$
The trouble for me seems to be in simplifying. I'm not the best at mathjax, so please forgive me for not typing out my work. I can kick it around until I get to something like
$$\sqrt{\frac{10+\sqrt10}{10}\over\frac{10-\sqrt10}{10}}$$
I've tried to to then reduce it to:
$$\sqrt{{10+\sqrt10}\over{10-\sqrt10}}$$ But from here on, multiplying the top and bottom by the conjugate isn't working. I think I just need a gentle push in the right direction. Thanks for any help!
UPDATE:
I spoke with my professor today and he pointed out that for the half-angle $\frac{\theta}{2}$ I had to remember to multiply the interval of the angle by $\frac{1}{2}$. So, instead of $\tan\theta=3$ being in quadrant III, $\tan\left(\frac{3}{2}\right)$ should be between $\frac{\pi}{2}$ and $\frac{3\pi}{4}$. This means my value of cosine (in the denominator) should also be negative. (To be clear, cosine would also be negative in QIII, I'm simply pointing out that I should have halved my interval.) That said, thanks again to those who helped me with my algebra/simplification, as well as those who suggested other ways of thinking about the exercise. You were a tremendous help!
Why don't you use tangent Double-Angle Formulas (List) $$\tan(2y)=\frac{2 \tan y}{1-\tan^2y}$$
We don't need to bother about $\displaystyle\cos\theta,\sin\theta$ and their proper signs