I am having trouble finding reference angles with radians. For instance: #1) 33pi/4
#2) -23pi/6
How do I go about solving these?
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I am having trouble finding reference angles with radians. For instance: #1) 33pi/4
#2) -23pi/6
How do I go about solving these?
Quote:
Originally Posted by davecolombia
Please do not be insulted by this remark: often it is a problem in basic arithmetic.
$\displaystyle \frac{33\pi}{4}-\frac{23\pi}{6}=\frac{99\pi}{12}-\frac{46\pi}{12}=\frac{53\pi}{12}=4\pi+\frac{5\pi} {12}$.
Now we can disregard all even multiples of $\displaystyle 2\pi$.
Thus in this case we endup with $\displaystyle \frac{5\pi}{12}$.
What is the reference angle there?
I apologize, I have written them wrong...they are separate questions: #1. 33pi/4 #2. -23pi/6
I apologize, I have written them wrong...they are separate questions: #1. 33pi/4 #2. -23pi/6
The fact that your post is mistaken makes no difference.Quote:
Originally Posted by davecolombia
We can disregard any multiple of $\displaystyle 2\pi$
Here is an example.
$\displaystyle \frac{76\pi}{5}=15\pi+\frac{\pi}{5}=14\pi+\frac{6 \pi}{5}$.
So disregard the $\displaystyle 14\pi$ and get the equivalent number of $\displaystyle \frac{6\pi}{5}$ .
Now what is the reference angle angle?
4pi/5 ?
How did you get that?
pi-pi/5=4pi/5
whoaa, wait..it is pi/5 which is 36 degrees ... right ?
5pi/12, which is 75 degrees
Here are the rules for reference angles.Quote:
Originally Posted by davecolombia
First reduce to $\displaystyle 0\le \theta <2\pi.$
The the reference angle is $\displaystyle \rho$ where
If $\displaystyle 0<\theta<\tfrac{\pi}{2}$ then $\displaystyle \rho=\theta$
If $\displaystyle \tfrac{\pi}{2}<\theta<\pi$ then $\displaystyle \rho=\pi-\theta$
If $\displaystyle \pi<\theta<\tfrac{3\pi}{2}$ then $\displaystyle \rho=\theta-\pi$
If $\displaystyle \tfrac{3\pi}{2}<\theta<2\pi$ then $\displaystyle \rho=2\pi-\theta$.
