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:


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


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!


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!

  • $\begingroup$ 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 $\sqrt{90} = 3\sqrt{10}$. $\endgroup$ – ShreevatsaR Apr 20 '14 at 15:52
  • $\begingroup$ Thanks, @ShreevatsaR! Somehow I reduce this to $$\frac{\sqrt{11-2{\sqrt{10}}}}{3}$$which appears to be wrong. I'll keep trying! $\endgroup$ – Jason S. Apr 20 '14 at 16:24
  • $\begingroup$ 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. $\endgroup$ – ShreevatsaR Apr 20 '14 at 16:49
  • $\begingroup$ @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. $\endgroup$ – Jason S. Apr 20 '14 at 20:20
  • $\begingroup$ I remember when I was new, I asked a similar question and had got 5 downvotes. $\endgroup$ – ibuprofen 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

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – Jason S. Apr 20 '14 at 15:34
  • $\begingroup$ (I'm new to this and still trying to develop a more intuitive understanding of how and when to apply these formulas.) $\endgroup$ – Jason S. Apr 20 '14 at 15:40
  • $\begingroup$ @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. $\endgroup$ – ShreevatsaR Apr 20 '14 at 15:46
  • $\begingroup$ @JasonS., We have $$\frac{2 \tan y}{1-\tan^2y}=\tan(2y)=3$$ and $$2y=\theta$$ $\endgroup$ – lab bhattacharjee Apr 20 '14 at 15:48
  • $\begingroup$ Thanks to everyone for your help! I'm going to try to use these tips and unravel the knots I've tied. I'll report back with my results and try to post my work for others similarly confused. $\endgroup$ – Jason S. Apr 20 '14 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.