# Hint finding exact value of half-angle when $\tan (\theta) = {3}$

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!

• You are doing right so far: your $$\sqrt{{10+\sqrt{10}}\over{10-\sqrt{10}}}$$ is equal to (by your idea of "multiplying the top and bottom by the conjugate") $$\sqrt{\frac{10 + \sqrt{10}}{10 - \sqrt{10}}\frac{10+\sqrt{10}}{10 + \sqrt{10}}}$$ If you get stuck again, note that $\sqrt{90} = 3\sqrt{10}$. – ShreevatsaR Apr 20 '14 at 15:52
• Thanks, @ShreevatsaR! Somehow I reduce this to $$\frac{\sqrt{11-2{\sqrt{10}}}}{3}$$which appears to be wrong. I'll keep trying! – Jason S. Apr 20 '14 at 16:24
• In the fraction under the square root (in my comment above), the numerator $(10 + \sqrt{10})(10 + \sqrt{10})$ is of course $(10 + \sqrt{10})^2$ (you can leave it like that, because you're going to take its square root next). The denominator $(10 - \sqrt{10})(10 + \sqrt{10})$ is $10^2 - (\sqrt{10})^2 = 100 - 10 = 90$. You've made some mistake somewhere in your calculation. – ShreevatsaR Apr 20 '14 at 16:49
• @ShreevatsaR: I tried to post a reply, but my mathjax code was full of errors. I feel silly for missing that I could have left the numerator squared. I'm still working on this. I managed to get it down to $\frac{\sqrt{10}+1}{3}$, but I missed the negative in my denominator somewhere. – Jason S. Apr 20 '14 at 20:20
• I remember when I was new, I asked a similar question and had got 5 downvotes. – MathDude3013 Sep 20 '18 at 11:49

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
• Great suggestion! I tried that previously plugging it in like this, but that gave me this: $$\frac{2(3)}{1-(3)^2}$$which yields $$-\frac{3}{4}$$Since that is equal to $2\theta$ which is twice angle, should I then plug that value into the half-angle formula to find the correct value? – Jason S. Apr 20 '14 at 15:34
• @JasonS.: the formula gives $\tan(2y)$ in terms of $\tan y$. What you've done is find $\tan(2\theta)$, instead of finding $\tan(\theta/2)$ as the question asks. Instead, try letting $y = \theta/2$, and apply the formula and see what turns up. – ShreevatsaR Apr 20 '14 at 15:46
• @JasonS., We have $$\frac{2 \tan y}{1-\tan^2y}=\tan(2y)=3$$ and $$2y=\theta$$ – lab bhattacharjee Apr 20 '14 at 15:48