# Math Help - all solutions of equation involving tan

1. ## all solutions of equation involving tan

Hi,

Let $0 \leq \theta \leq \pi$

Find all solutions to:

$\tan(\theta + \dfrac{\pi}{6})=\tan(\pi-\theta)$

Equating the two arguements and solving we find $\theta = \dfrac{5\pi}{12}$. How do I find the other solution?

Thanks

2. ## Re: all solutions of equation involving tan

Originally Posted by Ant
Hi,

Let $0 \leq \theta \leq \pi$

Find all solutions to:

$\tan(\theta + \dfrac{\pi}{6})=\tan(\pi-\theta)$

Equating the two arguements and solving we find $\theta = \dfrac{5\pi}{12}$. How do I find the other solution?

Thanks
Recall that tangent is periodic over 0 to Pi...

-Dan

3. ## Re: all solutions of equation involving tan

Here is how I would do it:

$\tan\left(\theta + \dfrac{\pi}{6} \right) = \dfrac{\sin\left(\theta + \dfrac{\pi}{6}\right)}{\cos \left( \theta + \dfrac{\pi}{6} \right)}$

and

$\tan\left(\pi - \theta \right) = \dfrac{\sin\left(\pi - \theta\right)}{\cos \left( \pi - \theta \right)}$

Cross multiplying gives:

$\sin\left(\theta + \dfrac{\pi}{6}\right)\cos\left(\pi - \theta\right) = \cos\left(\theta + \dfrac{\pi}{6}\right)\sin\left(\pi - \theta\right)$

Pulling everything to the LHS, you get:

$\sin\left(\theta + \dfrac{\pi}{6}\right)\cos\left(\pi - \theta\right) - \cos\left(\theta + \dfrac{\pi}{6}\right)\sin\left(\pi - \theta\right) = 0$

The LHS is just the difference of angles formula for sine. So you have:

$\sin\left(2\theta - \dfrac{5\pi}{6} \right) = 0$

You should know that sine equals zero at $n\pi$ for any integer $n$. So, $2\theta - \dfrac{5\pi}{6} = n\pi$. Hence, $\theta = \dfrac{\pi(6n+5)}{12}$.

As topsquark pointed out, tangent has a period of $\pi$, but it did not seem obvious to me that meant you will find solutions every $\dfrac{\pi}{2}$.

Edit: Now that I think about it, topsquark's hint was more straightforward than I thought. You set $\theta + \dfrac{\pi}{6} = \pi-\theta$. Instead, you needed to add $n\pi$ to one side. For example: $\theta + \dfrac{\pi}{6} = \pi-\theta + n\pi$. That would give you the same answer I got using the difference of angles formula for sine.

4. ## Re: all solutions of equation involving tan

Originally Posted by Ant
Hi,

Let $0 \leq \theta \leq \pi$

Find all solutions to:

$\tan(\theta + \dfrac{\pi}{6})=\tan(\pi-\theta)$

Equating the two arguements and solving we find $\theta = \dfrac{5\pi}{12}$. How do I find the other solution?

Thanks
First of all, remember that \displaystyle \begin{align*} \tan{ \left( \alpha + \beta \right) } \equiv \frac{\tan{( \alpha )} + \tan{ ( \beta ) }}{ 1 - \tan{( \alpha )} \tan{ (\beta )}} \end{align*} and also \displaystyle \begin{align*} \tan{ ( \pi - \theta )} \equiv -\tan{(\theta ) } \end{align*}, so that means

\displaystyle \begin{align*} \tan{ \left( \theta + \frac{\pi}{6} \right) } &= \tan{ \left( \pi - \theta \right) } \\ \frac{\tan{ \left( \theta \right)} + \tan{ \left( \frac{\pi}{6} \right) } }{1 - \tan{ \left( \theta \right) } \tan{ \left( \frac{\pi}{6} \right) } } &= -\tan{ (\theta )} \\ \frac{\tan{( \theta )} + \frac{1}{\sqrt{3}} }{ 1 - \frac{1}{\sqrt{3}} \tan{ (\theta )} } &= -\tan{ (\theta )} \\ \frac{\frac{\sqrt{3}\tan{( \theta )} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - \tan{(\theta)} }{\sqrt{3}}} &= -\tan{(\theta)} \\ \frac{\sqrt{3}\tan{(\theta)} + 1}{\sqrt{3} - \tan{(\theta)} } &= -\tan{(\theta)} \\ \sqrt{3}\tan{(\theta)} + 1 &= -\tan{(\theta)} \left[ \sqrt{3} - \tan{(\theta)} \right] \\ \sqrt{3} \tan{(\theta)} + 1 &= \tan^2{(\theta)} - \sqrt{3}\tan{(\theta)} \\ 0 &= \tan^2{(\theta)} - 2\sqrt{3}\tan{(\theta)} - 1 \\ 0 &= \tan^2{(\theta)} - 2\sqrt{3}\tan{(\theta)} + \left( -\sqrt{3} \right) ^2 - \left( -\sqrt{3} \right) ^2 - 1 \\ 0 &= \left[ \tan{(\theta)} - \sqrt{3} \right] ^2 - 3 - 1 \\ 0 &= \left[ \tan{(\theta)} - \sqrt{3} \right] ^2 - 4 \\ 4 &= \left[ \tan{(\theta)} - \sqrt{3} \right] ^2 \\ \pm 2 &= \tan{(\theta)} - \sqrt{3} \\ \sqrt{3} \pm 2 &= \tan{(\theta)} \\ \theta &= \arctan{ \left( \sqrt{3} \pm 2 \right) } + \pi n \textrm{ where } n \in \mathbf{Z} \end{align*}