Understanding the Question: csc^-1 (cotanΘ)

  • Thread starter Thread starter lucifer_x
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves finding csc^-1 (cotanΘ) given that tan^-1 (3/5) = Θ. Participants express confusion regarding the question and explore various approaches to clarify the relationship between the trigonometric functions involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants suggest rewriting the problem in terms of sines and cosines. Others propose drawing a right triangle to visualize the relationships between the sides and angles, particularly focusing on the tangent function. There is also mention of using a calculator as a straightforward method to find the necessary values.

Discussion Status

The discussion is active, with participants offering different methods to approach the problem. While some guidance has been provided, there is no explicit consensus on the best method to proceed. Various interpretations of the problem are being explored.

Contextual Notes

Participants are working under the assumption that they need to derive cotanΘ from the given tangent value, and there is an emphasis on understanding the relationships between the trigonometric functions involved.

lucifer_x
Messages
15
Reaction score
0

Homework Statement



If tan^-1 (3/5) = Θ , what is csc^-1 (cotanΘ)

i don't get the question
 
Physics news on Phys.org
Try writing everything in terms of sines and cosines.
 
Another way to do this is to draw a triangle. Since tangent= "opposite leg over near leg", draw a right triangle havine legs of length 3 and 4, with the leg of length 3 opposite angle [itex]\theta[/itex]. Find the length of the hypotenuse using the the Pythagorean theorem (if it isn't obvious) and then it should be simple to find cotan([itex]\theta[/itex]).

Another, rather obvious, way to do this is to use a calculator!
 
tan^-1 (3/5) = Θ , what is csc^-1 (cotanΘ)

If tan^-1(3/5) = Θ then tan(Θ) = 3/5

csc^-1(cot(Θ)) = csc^-1(1/tan(Θ)) = csc^-1(1/3/5) = csc^-1(5/3)

You should be able to take it from there.
 

Similar threads

Replies
24
Views
3K
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
10
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
18
Views
3K