# Thread: I am definitely stuck!

1. ## I am definitely stuck!

so here is my problem

tan theta + tan(theta +120degrees) + tan(theta + 240 degrees) = 3tan(3theta)

and here is what the answer is supposed to be......

tan + cot(pi/6-theta) - cot(theta + pi/6) = 3tan(3theta)

i get pretty far in my simplification process, what I don't understand is how does 120 degrees and 240 degrees each become pi/6? This is probably something really simple but if someone could please take pity on me and explain it, I would be most grateful!

sincerely
Nikki

2. ## Re: I am definitely stuck!

Originally Posted by ansulli2
so here is my problem

tan theta + tan(theta +120degrees) + tan(theta + 240 degrees) = 3tan(3theta)

and here is what the answer is supposed to be......

tan + cot(pi/6-theta) - cot(theta + pi/6) = 3tan(3theta)

i get pretty far in my simplification process, what I don't understand is how does 120 degrees and 240 degrees each become pi/6? This is probably something really simple but if someone could please take pity on me and explain it, I would be most grateful!

sincerely
Nikki
What, exactly, are you trying to solve for here? State the problem completely as it appears in the text. Your answer makes no sense. Your first tan function has no argument. Try again.

3. ## Re: I am definitely stuck!

I am supposed to establish the identity it says. The top equation i posted is the original problem. I am supposed to prove why it is correct.

4. ## Re: I am definitely stuck!

Originally Posted by ansulli2
so here is my problem

tan theta + tan(theta +120degrees) + tan(theta + 240 degrees) = 3tan(3theta)

and here is what the answer is supposed to be......

tan + cot(pi/6-theta) - cot(theta + pi/6) = 3tan(3theta)

i get pretty far in my simplification process, what I don't understand is how does 120 degrees and 240 degrees each become pi/6? This is probably something really simple but if someone could please take pity on me and explain it, I would be most grateful!

sincerely
Nikki
\displaystyle \begin{align*} 120^{\circ} = \frac{2\pi}{3} = \pi - \frac{\pi}{3} \end{align*} and \displaystyle \begin{align*} 240^{\circ} = \frac{4\pi}{3} = \pi + \frac{\pi}{3}\end{align*}, so you should be able to simplify using symmetry identities.

5. ## Re: I am definitely stuck!

I think the pi/6 is supposed to be 2pi/3 and 4pi/3 respectively

That would make more sense to me.

6. ## Re: I am definitely stuck!

Originally Posted by ansulli2
I think the pi/6 is supposed to be 2pi/3 and 4pi/3 respectively

That would make more sense to me.
Another hint:

\displaystyle \begin{align*} \sin{\left(\frac{\pi}{2} - \theta\right)} &\equiv \cos{\left(\theta\right)} \end{align*} and \displaystyle \begin{align*} \cos{\left(\frac{\pi}{2} - \theta\right)} &\equiv \sin{\left(\theta\right)} \end{align*}

Therefore:

\displaystyle \begin{align*} \tan{\left(\frac{\pi}{2} - \theta\right)} &\equiv \frac{\sin{\left(\frac{\pi}{2} - \theta\right)}}{\cos{\left(\frac{\pi}{2} - \theta\right)}} \\ &\equiv \frac{\cos{\left(\theta\right)}}{\sin{\left(\theta \right)}} \\ &\equiv \cot{\left(\theta\right)} \end{align*}