# Thread: A little help with reference angles.

1. ## A little help with reference angles.

So this is from a calculus problem but I'm not sure how to achieve the other angle (240 degrees). I know you get the (120 degrees from reference angle (180-60 = 120) but for 240 I'm not sure. Thanks and heres the problem.

$\displaystyle 2\cos t+1=0\Longrightarrow \cos t = -\frac{1}{2}\Longrightarrow t=\left\{\begin{array}{c}\frac{2\pi}{3}\\{}\\\frac {4\pi}{3}\end{array}\right.\,+2\pi k\,,\,\,k\in\mathbb{Z}$

How would I go about achieving the $\displaystyle \frac {4 \pi} 3$ ?
Thanks.

2. Originally Posted by elitenewbz
so this is from a calculus problem but i'm not sure how to achieve the other angle (240 degrees). I know you get the (120 degrees from reference angle (180-60 = 120) but for 240 i'm not sure. Thanks and heres the problem.

$\displaystyle 2\cos t+1=0\longrightarrow \cos t = -\frac{1}{2}\longrightarrow$$\displaystyle t=\left\{\begin{array}{c}\frac{2\pi}{3}\\{}\\\frac {4\pi}{3}\end{array}\right.\,+2\pi k\,,\,\,k\in\mathbb{z}$

how would i go about achieving the $\displaystyle \frac {4 \pi} 3$ ?
Thanks.
$\displaystyle 180+60=240$

3. So does that mean 240 = coterminal with the 60 degrees? I'm not sure how it fits... Von, which equation is that?

4. Originally Posted by EliteNewbz
So does that mean 240 = coterminal with the 60 degrees? I'm not sure how it fits... Von, which equation is that?
I'm not quite sure what you are asking me. It looks like you have done a great job at finding the solution in you first post.

I was only pointing out that $\displaystyle \cos{t}=\frac{-1}{2}$ implies that $\displaystyle t$ lies in $\displaystyle QII$ or $\displaystyle QIII$. So that this means that we have the two solutions

1. The one that you pointed out, namely $\displaystyle t=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

And

2. $\displaystyle \pi+\frac{\pi}{3}=\frac{4\pi}{3}$

for $\displaystyle t\in[0,2\pi]$

But, you even GENERALIZED the solution to include ALL values of t. So, what's your question?

5. Oh I get it now. You've answered it. You've added $\displaystyle \frac \pi 3$ to $\displaystyle \pi$ to include it into Quadrant III correct? (The pi/3 comes from the cos(t) = -1/2). So lets say the answers are in QUAD 4, you would add $\displaystyle \frac {5 \pi} 6$ to $\displaystyle \frac \pi 3$. Correct?

6. Originally Posted by EliteNewbz
Oh I get it now. You've answered it. You've added $\displaystyle \frac \pi 3$ to $\displaystyle \pi$ to include it into Quadrant III correct? (The pi/3 comes from the cos(t) = -1/2). So lets say the answers are in QUAD 4, you would add $\displaystyle \frac {5 \pi} 6$ to $\displaystyle \frac \pi 3$. Correct?
If $\displaystyle \cos\theta$ is negative, then there will be 2 solutions (excluding $\displaystyle \theta=-1$). They are given by $\displaystyle \theta=\pi+\varphi$ and $\displaystyle \theta=\pi-\varphi$, where $\displaystyle \varphi$ is a refrence angle.

Likewise, if $\displaystyle \cos\theta$ is positive then the solutions will be given by $\displaystyle \theta=\varphi$ and $\displaystyle 2\pi-\varphi$.