# Given $\tan\theta=7/9$ and $\cos\theta<0$, find $\tan(\theta/2)$.

Given $\tan\theta=7/9$ and $\cos\theta<0$, find $\tan(\theta/2)$.

Since the angle is lying on quadrant 3, I got $$\cos\theta=-9/\sqrt{130}$$ and substitute it in $$\tan(\theta/2)=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$ However, there is a negative sign in the final answer $$\tan(\theta/2)=-\sqrt{\frac{1+\frac{9}{\sqrt{130}}}{1-\frac{9}{\sqrt{130}}}}$$ I know when we square root a number, there will have a plus or minus sign outside the square root, but I don't know why it pick the negative one in this case.

• because you know $\theta/2$ should be lying in quadrant 2 – Nick Oct 2 '16 at 10:53
• Oh yes, I have made a mistake on here XD I mixed up $$(\pi+(\theta/2)) \text{ and }(\theta/2)$$ Thank you very much! – Lam Oct 2 '16 at 10:56

Do you have a straightedge and compasses? If so: draw an angle in a coordinate plane with vertex at the origin, one leg along the positive $x$ axis and the other leg in the third quadrant. Then construct the bisecting line, drawing it in both directions from the origin. What is the slope of the line; what quadrants does it pass through?

You infer the sign from the quadrant of $\theta$. For $\theta$ in the third quadrant, $\theta/2$ is in the second or fourth quadrant (which we see in the exercise above), so its tangent must be negative.

Using $t$-formula

Let $t=\tan \dfrac{\theta}{2}$,

\begin{align*} \tan \theta &= \frac{2t}{1-t^2} \\ \frac{2t}{1-t^2} &= \frac{7}{9} \\ 7t^2+18t-7 &= 0 \\ t &= \frac{-9\pm \sqrt{130}}{7} \\ \cos \theta &= \frac{1-t^2}{1+t^2} \\ \frac{1-t^2}{1+t^2} &< 0 \\ t^2 &> 1 \end{align*}

Rejecting $t = \dfrac{-9+\sqrt{130}}{7}$, we have

$$\tan \frac{\theta}{2}=\frac{-9-\sqrt{130}}{7}$$

$$\tan\left(\theta\right)=\frac{7}{9}\Longleftrightarrow\theta=n\pi+\arctan\left(\frac{7}{9}\right)$$ Where $n\in\mathbb{Z}$
When we substiute that into $\cos\left(\theta\right)$: $$\cos\left(n\pi+\arctan\left(\frac{7}{9}\right)\right)=\begin{cases} \frac{9}{\sqrt{130}}\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space n\space\text{is}\space\text{even}\\ -\frac{9}{\sqrt{130}}\space\space\space\space\space\space\space\space\space\space\text{when}\space n\space\text{is}\space\text{odd} \end{cases}$$
When we substiute that into $\sin\left(\theta\right)$: $$\sin\left(n\pi+\arctan\left(\frac{7}{9}\right)\right)=\begin{cases} \frac{7}{\sqrt{130}}\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space n\space\text{is}\space\text{even}\\ -\frac{7}{\sqrt{130}}\space\space\space\space\space\space\space\space\space\space\text{when}\space n\space\text{is}\space\text{odd} \end{cases}$$
So, when $x\in\mathbb{R}$:
$$\tan\left(\frac{\theta}{2}\right)=\frac{\sin\left(\theta\right)}{1+\cos\left(\theta\right)}=\frac{\sin\left(n\pi+\arctan\left(\frac{7}{9}\right)\right)}{1+\cos\left(n\pi+\arctan\left(\frac{7}{9}\right)\right)}=\begin{cases} \frac{\sqrt{130}-9}{7}\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space n\space\text{is}\space\text{even}\\ -\frac{9+\sqrt{130}}{7}\space\space\space\space\space\space\space\space\space\space\text{when}\space n\space\text{is}\space\text{odd} \end{cases}$$