# Finding all possible solutions between -pi <= theta <= pi for trig functions

• March 15th 2013, 08:05 PM
camjerlams
Finding all possible solutions between -pi <= theta <= pi for trig functions
I have just been solving for all possible angles between zero and 2pi using the knowledge that tan(theta)=tan(pi+theta) or sin(theta)=sin(pi-theta) etc. But now I can't seem to solve for angles between -pi and pi. Surely theres something easy I'm missing, can someone help?
• March 15th 2013, 08:20 PM
Prove It
Re: Finding all possible solutions between -pi <= theta <= pi for trig functions
Quote:

Originally Posted by camjerlams
I have just been solving for all possible angles between zero and 2pi using the knowledge that tan(theta)=tan(pi+theta) or sin(theta)=sin(pi-theta) etc. But now I can't seem to solve for angles between -pi and pi. Surely theres something easy I'm missing, can someone help?

What's the actual equation you are trying to solve?
• March 15th 2013, 08:32 PM
camjerlams
Re: Finding all possible solutions between -pi <= theta <= pi for trig functions
Well there's several all of the same question, either with sinx, cosx or tanx.

Here's one:
tan(theta)=2.4 Find all possible solutions for theta, -pi <= theta <= pi

So I figured tan^-1(2.4) = 1.18 so this is a solution in the first quadrant, but you can also find this angle in the third quadrant for tan.

I suppose I need to find the corresponding theta in the third Q, normally I'd use tan(theta) = tan(pi+theta) to get it but this value is greater than pi so I thought to use tan(theta) = tan(theta-pi) but that's not right either neither is tan(-pi+theta) so im out of ideas
• March 15th 2013, 08:46 PM
Prove It
Re: Finding all possible solutions between -pi <= theta <= pi for trig functions
OK, well generally speaking, to solve a trigonometric equation, you remember that in one cycle of the unit circle there will be two solutions, and then those two solutions will be repeated after every angle of $\displaystyle 2\pi$. So in your specific question, where $\displaystyle \tan{(\theta)} = 2.4$, we notice that the tangent function will be positive in the first and third quadrants. The focus angle will be the angle in the first quadrant, and the other angle will be separated from the focus angle by $\displaystyle \pi$. Once you have these two values, you add all possible multiples of $\displaystyle 2\pi$.

\displaystyle \begin{align*} \tan{(\theta)} &= 2.4 \\ \theta &= \left\{ \tan^{-1}{(2.4)} , \pi + \tan^{-1}{(2.4)} \right\} + 2\pi\, n \textrm{ where } n \textrm{ takes on all possible integer values} \end{align*}

Since you are told a region that $\displaystyle \theta$ has to lie in, you now let n take on specific integer values to determine which ones lie in your region.

You should find that when n = 0, $\displaystyle \theta = \left\{ \tan^{-1}{(2.4)}, \pi + \tan^{-1}{(2.4)} \right\}$. The first one is in the region $\displaystyle -\pi \leq \theta \leq \pi$, but the second isn't. So any value of n bigger than 0 won't give acceptable answers.

If n = -1, we have

\begin{align*} \theta &= \left\{ \tan^{-1}{(2.4)} , \pi + \tan^{-1}{(2.4)} \right\} - 2\pi \\ &= \left\{ \tan^{-1}{(2.4)} - 2\pi, \tan^{-1}{(2.4)} - \pi \right\} \end{align*}

The first of these does not lie in the region $\displaystyle -\pi \leq \theta \leq \pi$, but the second does. So that means any value for n less than -1 won't give acceptable answer.

Therefore the solutions in your region are $\displaystyle \theta = \left\{ \tan^{-1}{(2.4)} - \pi , \tan^{-1}{(2.4)} \right\}$.