Trigonometric functions and the unit circle

Click For Summary

Homework Help Overview

The discussion revolves around solving the equation \(\cos \theta = -1/2\) and expressing the solutions in a general form involving an integer \(n\). The subject area includes trigonometric functions and the unit circle.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore how to express the specific solutions \(\theta = 2\pi/3\) and \(\theta = 4\pi/3\) in a general form. Questions arise regarding the feasibility of combining these solutions into a single expression involving an integer \(n\).

Discussion Status

Some participants suggest various forms for expressing the solutions, while others express uncertainty about the best way to combine them. There is acknowledgment of the common expression for the solutions, but no consensus on a single, neat formulation.

Contextual Notes

Participants are considering the implications of expressing trigonometric solutions in terms of integers and the limitations of doing so without a calculator or reference materials.

Niles
Messages
1,834
Reaction score
0

Homework Statement


Hi all.

Today I had to solve: [itex]\cos \theta = -1/2[/itex]. What I did was to look in a table to find that [itex]\theta = 2\pi/3 \quad \text{and}\quad \theta = 4\pi/3[/itex].

My question is what is the general strategy when I wish to write this as a a function of an integer n? Is there even a general strategy for this?

Thanks in advance.
 
Physics news on Phys.org
Niles said:
... when I wish to write this as a a function of an integer n? Is there even a general strategy for this?

What do you mean?
Can you solve
[tex]\cos\theta = n[/tex]
without looking it up (no, unless you have a calculator), or can you write the solutions
[tex]\theta = 2\pi/3, 4 \pi / 3[/tex]
as
[tex]\theta = \frac{2\pi}{3} + 2 \pi n, \frac{4 \pi}{3} + 2 \pi n[/tex]
or can you combine those into one formula (no, not particularly nicely).
 
What I meant was your #2 suggestion: If there is any way to write the two solutions nicely for all n.

Hmm, ok then. I will just stick to writing the solutions as you did. Thanks.
 
OK, so maybe you can write something like
[tex]\frac{2 \epsilon \pi}{3} + 2 \pi n[/tex]
with [itex]n \in \mathbb{Z}, \epsilon = 1, 2[/itex]
or
[tex]\frac{k \pi}{3}[/tex]
with [itex]k \in \mathbb{Z}, k \equiv 2 \text{ or } 4 \, \operatorname{mod} 7[/itex].

But I think the expression from my earlier post is more common (in any case, you only need to think about it half as long to understand what it's saying and where it comes from :smile:).
 
Ok, thanks!
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
2
Views
3K